大家应该都会玩“锤子剪刀布”的游戏：
现给出两人的交锋记录，请统计双方的胜、平、负次数，并且给出双方分别出什么手势的胜算最大。
[编程题]锤子剪刀布 (20)
- 热度指数：31488 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
输入描述:
输入第1行给出正整数N（<=105），即双方交锋的次数。随后N行，每行给出一次交锋的信息，即甲、乙双方同时给出的的手势。C代表“锤子”、J代表“剪刀”、B代
表“布”，第1个字母代表甲方，第2个代表乙方，中间有1个空格。
输出描述:
输出第1、2行分别给出甲、乙的胜、平、负次数，数字间以1个空格分隔。第3行给出两个字母，分别代表甲、乙获胜次数最多的手势，中间有1个空格。如果解不唯
一，则输出按字母序最小的解。
示例1
输入
10<br/>C J<br/>J B<br/>C B<br/>B B<br/>B C<br/>C C<br/>C B<br/>J B<br/>B C<br/>J J
输出
5 3 2<br/>2 3 5<br/>B B
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
int count = 0;
int[] bifen = new int[3];
char[] A = new char[num];
char[] B = new char[num];
int A_C = 0, A_J = 0, A_B =0, B_C = 0, B_J = 0, B_B =0;
while (num-- > 0){
A[count] = scanner.next().charAt(0);
B[count] = scanner.next().charAt(0);
if (A[count] == B[count]){
bifen[1]++;
}else if (A[count] == 'C' && B[count] == 'J'){
bifen[0]++;
A_C++;
}else if (A[count] == 'C' && B[count] == 'B'){
bifen[2]++;
B_B++;
}else if (A[count] == 'J' && B[count] == 'C'){
bifen[2]++;
B_C++;
}else if (A[count] == 'J' && B[count] == 'B'){
bifen[0]++;
A_J++;
}else if (A[count] == 'B' && B[count] == 'C'){
bifen[0]++;
A_B++;
}else if(A[count] == 'B' && B[count] == 'J'){
bifen[2]++;
B_J++;
}
}
System.out.println(bifen[0]+ " " + bifen[1] + " " + bifen[2]);
System.out.println(bifen[2]+ " " + bifen[1] + " " + bifen[0]);
if (A_C == A_B && A_C == A_J || A_B >= A_C && A_B >= A_J)
System.out.print("B ");
else if (A_C >= A_B && A_C >= A_J)
System.out.print("C ");
else
System.out.print("J ");
if (B_C == B_B && B_C == A_J || B_B >= B_C && B_B >= B_J)
System.out.print("B");
else if (B_C >= B_B && B_C >= B_J)
System.out.print("C");
else
System.out.print("J");
}
}
发表于 2019-07-01 21:14:54
回复(0)
更多回答
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int n = sc.nextInt();
int v = 0;
int f = 0;
Map<Character, Integer> mapA = new TreeMap<>();
Map<Character, Integer> mapB = new TreeMap<>();
for (int i = 0; i < n; i++) {
char a = sc.next().charAt(0);
char b = sc.next().charAt(0);
int re = match(a, b);
if (re == 1) {
v++;
if (mapA.containsKey(a)) {
mapA.put(a, mapA.get(a) + 1);
} else {
mapA.put(a, 1);
}
} else if (re == -1) {
f++;
if (mapB.containsKey(b)) {
mapB.put(b, mapB.get(b) + 1);
} else {
mapB.put(b, 1);
}
}
}
System.out.println(v + " " + (n - v - f) + " " + f);
System.out.println(f + " " + (n - v - f) + " " + v);
int max = 0;
for (Map.Entry<Character, Integer> m : mapA.entrySet()) {
if (m.getValue() > max) {
max = m.getValue();
}
}
if (max == 0) {
System.out.print("B");
} else {
for (Map.Entry<Character, Integer> m : mapA.entrySet()) {
if (m.getValue() == max) {
System.out.print(m.getKey());
break;
}
}
}
max = 0;
System.out.print(" ");
for (Map.Entry<Character, Integer> m : mapB.entrySet()) {
if (m.getValue() > max) {
max = m.getValue();
}
}
if (max == 0) {
System.out.print("B");
} else {
for (Map.Entry<Character, Integer> m : mapB.entrySet()) {
if (m.getValue() == max) {
System.out.print(m.getKey());
break;
}
}
}
}
}
public static int match(char a, char b) {
if (a == b) {
return 0;
} else if ((a == 'C' && b == 'J') || (a == 'J' && b == 'B') || (a == 'B' && b == 'C')) {
return 1;
} else {
return -1;
}
}
}
编辑于 2020-03-10 21:27:53
回复(1)
python solution
win, lose, tie = 0, 0, 0 #甲的勝、負、平次數
Jia, Yi = [0, 0, 0], [0, 0, 0] # B C J count #甲乙勝時出的手勢記錄
num = int(input())
for i in range(num):
string = input()
if string == "C J":
win += 1
Jia[1] += 1
elif string == "J C":
lose += 1
Yi[1] += 1
elif string == "C B":
lose += 1
Yi[0] += 1
elif string == "B C":
win += 1
Jia[0] += 2
elif string == "B J":
lose += 1
Yi[2] += 1
elif string == "J B":
win += 1
Jia[2] += 1
else: # there is a tie
tie += 1
print(win, tie, lose)
print(lose, tie, win)
print("BCJ"[Jia.index(max(Jia))], "BCJ"[Yi.index(max(Yi))])
发表于 2017-11-17 14:22:40
回复(0)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
private static int getMaxIndex(int[] arr) {
int maximumIndex = 0;
for(int i = 0; i < arr.length; i++) {
if(arr[i] > arr[maximumIndex]) {
maximumIndex = i;
}
}
return maximumIndex;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(br.readLine());
final String[] gesture = {"B" ,"C", "J"};
int[] ATime = {0, 0, 0};
int[] BTime = {0, 0, 0};
int[] AWinGesture = {0, 0, 0};
int[] BWinGesture = {0, 0, 0};
String[] game;
for(int i = 0; i < N; i++) {
game = br.readLine().split(" ");
switch(game[0]) {
case "C" :{
switch (game[1]) {
case "C":{
ATime[1]++;
BTime[1]++;
break;
}
case "J":{
ATime[0]++;
BTime[2]++;
AWinGesture[1]++;
break;
}
case "B":{
ATime[2]++;
BTime[0]++;
BWinGesture[0]++;
break;
}
default:break;
}
break;
}
case "J" :{
switch (game[1]) {
case "C":{
ATime[2]++;
BTime[0]++;
BWinGesture[1]++;
break;
}
case "J":{
ATime[1]++;
BTime[1]++;
break;
}
case "B":{
ATime[0]++;
BTime[2]++;
AWinGesture[2]++;
break;
}
default:break;
}
break;
}
case "B" :{
switch (game[1]) {
case "C":{
ATime[0]++;
BTime[2]++;
AWinGesture[0]++;
break;
}
case "J":{
ATime[2]++;
BTime[0]++;
BWinGesture[2]++;
break;
}
case "B":{
ATime[1]++;
BTime[1]++;
break;
}
default:break;
}
break;
}
default : break;
}
}
System.out.println(ATime[0] + " " + ATime[1] + " " + ATime[2]);
System.out.println(BTime[0] + " " + BTime[1] + " " + BTime[2]);
System.out.println(gesture[getMaxIndex(AWinGesture)] + " " + gesture[getMaxIndex(BWinGesture)]);
}
}
发表于 2017-11-15 18:07:54
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
int[] js = new int[3];
int jf = 0;
int[] ys = new int[3];
int yf = 0;
int p = 0;
while(N--!=0){
int j = convert(in.next());
int y = convert(in.next());
if(j==y)
p++;
else if(judge(j, y)){
//甲赢
js[j]++;
yf++;
}else{
ys[y]++;
jf++;
}
}
System.out.println(yf+" "+p+" "+jf);
System.out.println(jf+" "+p+" "+yf);
System.out.println(getResult(js)+" "+getResult(ys));
}
private static int convert(String s){
if(s.equals("B"))
return 0;
if(s.equals("C"))
return 1;
return 2;// J
}
private static boolean judge(int j,int y){
if(j==2&&y==0)
return true;
else if(j>y)
return false;
return true;
}
private static String getResult(int[] a){
int max = a[0];
int index = 0;
for(int i = 1;i<a.length;i++){
if(a[i]>max){
max = a[i];
index = i;
}
}
return index==0?"B":index==1?"C":"J";
}
}
发表于 2016-06-08 21:54:32
回复(0)
//记为ab两人 a胜利的情况为：CJ、BC、JB。b胜利的情况为：JC、CB、BJ。
//根据两人胜利的情况分别统计出两人胜利的次数。平局次数 = 总场数-a胜利-b胜利。
// 用LinkedHashMap记录下ab胜利使用次数的情况。输出胜利次数最多使用情况，如果相同按字母顺序输出。
//会有三种极端情况
//情况一二：ab两人中一人全胜。举例：若a全胜且使用次数最多为C则此时b虽然全输，但是应该输出B
//因为题目描述为：双方分别出什么手势胜算最大。
//情况三：ab全平。此时胜算最大无输出，（牛客应该没有情况三的用例）
import java.util.*;
public class Main {
private static int winA = 0;
private static int winB = 0;
private static HashMap<Character,Integer> A = new LinkedHashMap<Character,Integer>();
private static HashMap<Character,Integer> B = new LinkedHashMap<Character,Integer>();
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()) {
int n = sc.nextInt();
int count = n;
sc.nextLine();
while(count-- > 0) {
String[] str = sc.nextLine().split(" ");
StringBuffer sb = new StringBuffer();
for(String s : str) {
sb.append(s);
}
Cjb(sb.toString());
}
int peace = n-winA-winB;
System.out.println(winA+" "+peace+" "+winB);
System.out.println(winB+" "+peace+" "+winA);
if(A.size() != 0 && B.size() != 0) {
System.out.println(Mvp(A)+" "+Mvp(B));
}
if(A.size() == 0 && B.size() != 0) {
char a = (Mvp(B) == 'B') ? 'J' :((Mvp(B) == 'C') ? 'B' : 'C');
System.out.println(a+" "+Mvp(B));
}
if(A.size() != 0 && B.size() == 0) {
char b = (Mvp(A) == 'B') ? 'J' :((Mvp(A) == 'C') ? 'B' : 'C');
System.out.println(Mvp(A)+" "+b);
}
}
}
private static char Mvp(HashMap<Character,Integer> map) {
Set<Character> set = map.keySet();
int tmp = 0;
char ret = 'Z';
for(char c : set) {
if(tmp <= map.get(c)) {
tmp = map.get(c);
if(ret > c) {
ret = c;
}
}
}
return ret;
}
private static void Cjb(String s) {
if(s.equals("CJ") || s.equals("JB") || s.equals("BC")) {
winA++;
A.put(s.charAt(0),A.containsKey(s.charAt(0)) ? A.get(s.charAt(0))+1 : 1);
}
if(s.equals("JC") || s.equals("BJ") || s.equals("CB")) {
winB++;
B.put(s.charAt(1),B.containsKey(s.charAt(1)) ? B.get(s.charAt(1))+1 : 1);
}
}
}
发表于 2020-04-08 15:01:56
回复(0)
和大家的思路差不多的亚子，容易理解的通俗解法
#include <iostream>
using namespace std;
/*
C > J
J > B
B > C
B > C > J > B
*/
int main(int argc, char** argv) {
int num;
// 0:胜， 1:平， 2:负， 3:B， 4:C， 5:J
int jarr[6] = {0}, yarr[6] = {0};
char ja, yi;
cin >> num;
while(num--){ // 求获胜次数获胜字母的次数
cin >> ja >> yi;
if(ja == 'B' && yi == 'C'){
jarr[0]++;
yarr[2]++;
jarr[3]++;
}else if(ja == 'C' && yi == 'B'){
jarr[2]++;
yarr[0]++;
yarr[3]++;
}else if(ja == 'J' && yi == 'B'){
jarr[0]++;
yarr[2]++;
jarr[5]++;
}else if(ja == 'B' && yi == 'J'){
jarr[2]++;
yarr[0]++;
yarr[5]++;
}else if(ja == 'C' && yi == 'J'){
jarr[0]++;
yarr[2]++;
jarr[4]++;
}else if(ja == 'J' && yi == 'C'){
jarr[2]++;
yarr[0]++;
yarr[4]++;
}else{
jarr[1]++;
yarr[1]++;
}
}
ja = jarr[4] >= jarr[5] ? 'C' : 'J';
ja = jarr[3] >= jarr[4] ? 'B' : ja;
yi = yarr[4] >= yarr[5] ? 'C' : 'J';
yi = yarr[3] >= yarr[4] ? 'B' : yi;
for(int i = 0; i < 3; i++){
cout << jarr[i];
if(i!=2) cout << " ";
}
cout << endl;
for(int i = 0; i < 3; i++){
cout << yarr[i];
if(i!=2) cout << " ";
}
cout << endl;
cout << ja << " " << yi;
return 0;
}
发表于 2020-02-27 21:20:39
回复(0)
#include<iostream>
#include<string>
using namespace std;
struct node{
char x;
char y;
};
int main(){
int n;
cin>>n;
node s[106];
int sjcount=0,fjcount=0,p=0;
string strj,stry;
for(int i=0;i<n;i++){
cin>>s[i].x>>s[i].y;
if(s[i].x==s[i].y)p++;
else {
if(s[i].x=='C'){
if(s[i].y=='J'){
sjcount++;
strj.push_back('C');
}
else{
fjcount++;
stry.push_back('B');
}
}
else if(s[i].x=='B'){
if(s[i].y=='C'){
sjcount++;
strj.push_back('B');
}
else{
fjcount++;
stry.push_back('J');
}
}
else if(s[i].x=='J'){
if(s[i].y=='B'){
sjcount++;
strj.push_back('J');
}
else{
fjcount++;
stry.push_back('C');
}
}
}
}
int countjJ=0,countjB=0,countjC=0,countyJ=0,countyB=0,countyC=0;
for(int i=0;i<strj.length();i++){
if(strj[i]=='J')countjJ++;
else if(strj[i]=='B')countjB++;
else countjC++;
}
for(int i=0;i<stry.length();i++){
if(stry[i]=='J')countyJ++;
else if(stry[i]=='B')countyB++;
else countyC++;
}
cout<<sjcount<<" "<<p<<" "<<fjcount<<endl;
cout<<fjcount<<" "<<p<<" "<<sjcount<<endl;
if(countjB>=countjJ&&countjB>=countjC)cout<<"B"<<" ";
else if(countjC>countjB&&countjC>=countjJ)cout<<"C"<<" ";
else cout<<"J"<<" ";
if(countyB>=countyJ&&countyB>=countyC)cout<<"B";
else if(countyC>countyB&&countyC>=countyJ)cout<<"C";
else cout<<"J";
return 0;
}
#include<string>
using namespace std;
struct node{
char x;
char y;
};
int main(){
int n;
cin>>n;
node s[106];
int sjcount=0,fjcount=0,p=0;
string strj,stry;
for(int i=0;i<n;i++){
cin>>s[i].x>>s[i].y;
if(s[i].x==s[i].y)p++;
else {
if(s[i].x=='C'){
if(s[i].y=='J'){
sjcount++;
strj.push_back('C');
}
else{
fjcount++;
stry.push_back('B');
}
}
else if(s[i].x=='B'){
if(s[i].y=='C'){
sjcount++;
strj.push_back('B');
}
else{
fjcount++;
stry.push_back('J');
}
}
else if(s[i].x=='J'){
if(s[i].y=='B'){
sjcount++;
strj.push_back('J');
}
else{
fjcount++;
stry.push_back('C');
}
}
}
}
int countjJ=0,countjB=0,countjC=0,countyJ=0,countyB=0,countyC=0;
for(int i=0;i<strj.length();i++){
if(strj[i]=='J')countjJ++;
else if(strj[i]=='B')countjB++;
else countjC++;
}
for(int i=0;i<stry.length();i++){
if(stry[i]=='J')countyJ++;
else if(stry[i]=='B')countyB++;
else countyC++;
}
cout<<sjcount<<" "<<p<<" "<<fjcount<<endl;
cout<<fjcount<<" "<<p<<" "<<sjcount<<endl;
if(countjB>=countjJ&&countjB>=countjC)cout<<"B"<<" ";
else if(countjC>countjB&&countjC>=countjJ)cout<<"C"<<" ";
else cout<<"J"<<" ";
if(countyB>=countyJ&&countyB>=countyC)cout<<"B";
else if(countyC>countyB&&countyC>=countyJ)cout<<"C";
else cout<<"J";
return 0;
}
发表于 2020-02-23 12:52:16
回复(0)
//锤子剪刀布
int main()
{
int n;
cin >> n;
int p1 = 0, p2 = 0, p3 = 0;
int q1 = 0, q2 = 0, q3 = 0;
int equ = 0;
for (int i = 0; i < n; i++)
{
char a, b;
cin >> a >> b;
if (a == 'C'&&b == 'J') p1++;
else if (a == 'C'&&b == 'B') q3++;
else if (a == 'C'&&b == 'C') equ++;
else if (a == 'J'&&b == 'J') equ++;
else if (a == 'J'&&b == 'B') p2++;
else if (a == 'J'&&b == 'C') q1++;
else if (a == 'B'&&b == 'J') q2++;
else if (a == 'B'&&b == 'B') equ++;
else if (a == 'B'&&b == 'C') p3++;
}
int pmax = p3, qmax = q3;
if (pmax < p1) pmax = p2;
if (pmax < p2) pmax = p1;
if (qmax < q1) qmax = q2;
if (qmax < q2) qmax = q1;
cout << p1 + p2 + p3 << ' ' << equ << ' ' << q1 + q2 + q3 << endl;
cout << q1 + q2 + q3 << ' ' << equ << ' ' << p1 + p2 + p3 << endl;
if (pmax == p3)cout << 'B';
else if (pmax == p1)cout << 'C';
else cout << 'J';
cout << ' ';
if (qmax == q3)cout << 'B';
else if (qmax == q1)cout << 'C';
else cout << 'J';
cout << endl;
return 0;
}
发表于 2019-08-17 16:54:02
回复(0)
/*
甲胜利有3种情况
C J
J B
B C
乙胜利反之
J C
B J
C B
此题在输入时判断次数，则不需要开数组
求出甲胜利、平局的次数和甲、乙出锤、剪、布胜利的次数即可
*/
#include<iostream>
using namespace std;
char jia,yi;
int times;
int jiaWin=0,Ping=0;
int jiaChui=0,jiaBu=0,jiaJian=0;
int yiChui=0,yiBu=0,yiJian=0;
int i;
int main(){
cin>>times;
for(i=0;i<times;i++){
cin>>jia>>yi;
if(jia=='C'&&yi=='J'){
jiaWin++;
jiaChui++;
}else if(jia=='J'&&yi=='B'){
jiaWin++;
jiaJian++;
}else if(jia=='B'&&yi=='C'){
jiaWin++;
jiaBu++;
}else if(jia==yi)Ping++;
else if(yi=='C'&&jia=='J')yiChui++;
else if(yi=='J'&&jia=='B')yiJian++;
else if(yi=='B'&&jia=='C')yiBu++;
}
cout<<jiaWin<<" "<<Ping<<" "<<times-Ping-jiaWin<<endl
<<times-Ping-jiaWin<<" "<<Ping<<" "<<jiaWin<<endl;
if(jiaBu>=jiaChui&&jiaBu>=jiaJian)cout<<"B ";
else if(jiaChui>=jiaJian&&jiaChui>jiaBu)cout<<"C ";
else cout<<"J ";
if(yiBu>=yiChui&&yiBu>=yiJian)cout<<"B";
else if(yiChui>=yiJian&&yiChui>yiBu)cout<<"C";
else cout<<"J";
return 0;
}
编辑于 2019-07-24 12:36:15
回复(0)
#include <iostream>
using namespace std;
struct Caiquan
{
int jia_win=0;
int jia_lose=0;
int tie=0;
int yi_win=0;
int yi_lose=0;
int jia_J=0;
int jia_C=0;
int jia_B=0;
int yi_J=0;
int yi_C=0;
int yi_B=0;
}caiquan;
int main()
{
int N,i;
char j,y;
cin>>N;
for(i=0;i<N;i++)
{
cin>>j;
getchar();
cin>>y;
if((j=='J'&&y=='J')||(j=='C'&&y=='C')||(j=='B'&&y=='B'))
{
caiquan.tie+=1;
}
else if(j=='J'&&y=='B')
{
caiquan.jia_win+=1;
caiquan.yi_lose+=1;
caiquan.jia_J+=1;
}
else if(j=='C'&&y=='J')
{
caiquan.jia_win+=1;
caiquan.yi_lose+=1;
caiquan.jia_C+=1;
}
else if(j=='B'&&y=='C')
{
caiquan.jia_win+=1;
caiquan.yi_lose+=1;
caiquan.jia_B+=1;
}
else if(y=='J'&&j=='B')
{
caiquan.jia_lose+=1;
caiquan.yi_win+=1;
caiquan.yi_J+=1;
}
else if(y=='C'&&j=='J')
{
caiquan.jia_lose+=1;
caiquan.yi_win+=1;
caiquan.yi_C+=1;
}
else if(y=='B'&&j=='C')
{
caiquan.jia_lose+=1;
caiquan.yi_win+=1;
caiquan.yi_B+=1;
}
}
cout<<caiquan.jia_win<<" "<<caiquan.tie<<" "<<caiquan.jia_lose<<endl;
cout<<caiquan.yi_win<<" "<<caiquan.tie<<" "<<caiquan.yi_lose<<endl;
if(caiquan.jia_B>=caiquan.jia_C&&caiquan.jia_B>=caiquan.jia_J)cout<<"B ";
else if(caiquan.jia_C>caiquan.jia_B&&caiquan.jia_C>=caiquan.jia_J)cout<<"C ";
else if(caiquan.jia_J>caiquan.jia_B&&caiquan.jia_J>caiquan.jia_C)cout<<"J ";
if(caiquan.yi_B>=caiquan.yi_C&&caiquan.yi_B>=caiquan.yi_J)cout<<"B";
else if(caiquan.yi_C>caiquan.yi_B&&caiquan.yi_C>=caiquan.yi_J)cout<<"C";
else if(caiquan.yi_J>caiquan.yi_B&&caiquan.yi_J>caiquan.yi_C)cout<<"J";
}
发表于 2019-07-11 14:45:27
回复(0)
- #include <bits/stdc++.h>
- using namespace std;
- map<string,int>q1,q2;
- map<string,int>::iterator it;
- int main()
- {
- int n;
- cin>>n;
- string s1,s2;
- int a1,b1,c1;
- int a2,b2,c2;
- a1=a2=b1=b2=c1=c2=0;
- for(int i=0; i<n; i++)
- {
- cin>>s1>>s2;
- if(s1=="C"&&s2=="J"||s1=="J"&&s2=="B"||s1=="B"&&s2=="C")
- {
- a1++;
- c2++;
- q1[s1]++;
- }
- else if(s2=="C"&&s1=="J"||s2=="J"&&s1=="B"||s2=="B"&&s1=="C")
- {
- a2++;
- c1++;
- q2[s2]++;
- }
- else if(s1==s2)
- {
- b1++;
- b2++;
- q1[s1]++;
- q2[s1]++;
- }
- }
- printf("%d %d %d\n",a1,b1,c1);
- printf("%d %d %d\n",a2,b2,c2);
- int l,r;
- string l1,r1;
- l=r=0;
- for(it=q1.begin(); it!=q1.end(); it++)
- {
- if(it->second>l)
- {
- l1=it->first;
- l=it->second;
- }
- }
- for(it=q2.begin(); it!=q2.end(); it++)
- {
- if(it->second>r)
- {
- r1=it->first;
- r=it->second;
- }
- }
- if(a1!=0&&a2==0)
- {
- if(l1=="C")
- r1="B";
- if(l1=="J")
- r1="C";
- if(l1=="B")
- r1="J";
- }
- else if(a1==0&&a2!=0)
- {
- if(r1=="C")
- l1="B";
- if(r1=="J")
- l1="C";
- if(r1=="B")
- l1="J";
- }
- cout<<l1<<" "<<r1<<endl;
- return 0;
- }
编辑于 2018-12-13 23:46:02
回复(0)
#include<iostream>
using namespace std;
int main(){
int times;
while(cin>>times){
char a, b;
int awin = 0, bwin = 0, alose = 0, blose = 0, balance = 0;
int aJ = 0, aC = 0, aB = 0, bJ = 0, bC = 0, bB = 0;
while(times--){
cin>>a>>b;
if((a=='J'&&b=='B')||(a=='B'&&b=='C')||(a=='C'&&b=='J'))
{
awin++;
blose++;
if(a=='J')
{
aJ++;
}
else if(a=='B'){
aB++;
}
else{
aC++;
}
}
else if(a==b)
{
balance++;
}
else{
bwin++;
alose++;
if(b=='J')
{
bJ++;
}
else if(b=='B'){
bB++;
}
else{
bC++;
}
}
}
cout<<awin<<" "<<balance<<" "<<alose<<endl;
cout<<bwin<<" "<<balance<<" "<<blose<<endl;
if(aJ>aB)
{
if(aB>=aC)
cout<<'J'<<" ";
else if(aJ==aC)
cout<<'C'<<" ";
}
else if(aJ>=aC)
{
cout<<'B'<<" ";
}
else
{
cout<<'C'<<" ";
}
if(bJ>bB)
{
if(bB>=bC)
cout<<'J'<<endl;
else if(bJ==bC)
cout<<'C'<<endl;
}
else if(bJ>=bC)
{
cout<<'B'<<endl;
}
else
{
cout<<'C'<<endl;
}
}
return 0;
}
using namespace std;
int main(){
int times;
while(cin>>times){
char a, b;
int awin = 0, bwin = 0, alose = 0, blose = 0, balance = 0;
int aJ = 0, aC = 0, aB = 0, bJ = 0, bC = 0, bB = 0;
while(times--){
cin>>a>>b;
if((a=='J'&&b=='B')||(a=='B'&&b=='C')||(a=='C'&&b=='J'))
{
awin++;
blose++;
if(a=='J')
{
aJ++;
}
else if(a=='B'){
aB++;
}
else{
aC++;
}
}
else if(a==b)
{
balance++;
}
else{
bwin++;
alose++;
if(b=='J')
{
bJ++;
}
else if(b=='B'){
bB++;
}
else{
bC++;
}
}
}
cout<<awin<<" "<<balance<<" "<<alose<<endl;
cout<<bwin<<" "<<balance<<" "<<blose<<endl;
if(aJ>aB)
{
if(aB>=aC)
cout<<'J'<<" ";
else if(aJ==aC)
cout<<'C'<<" ";
}
else if(aJ>=aC)
{
cout<<'B'<<" ";
}
else
{
cout<<'C'<<" ";
}
if(bJ>bB)
{
if(bB>=bC)
cout<<'J'<<endl;
else if(bJ==bC)
cout<<'C'<<endl;
}
else if(bJ>=bC)
{
cout<<'B'<<endl;
}
else
{
cout<<'C'<<endl;
}
}
return 0;
}
发表于 2018-08-10 19:13:32
回复(0)
#include <bits/stdc++.h>
using namespace std;
int main()
{
int m,i,j,s[80]={0},s1[80]={0},t=0,t1=0,t2=0,a[2][3]={0},max=0,max1=0,a1,b;
char no,st;
cin >> m;
for(i = 0; i < m; i++)
{
cin >> no >>st;
if(no - st == -7 || no - st == 8 || no - st == -1)
{
a[0][0]++;
a[1][2]++;
if(no - st == -7)s[no]++;
if(no - st == 8)s[no]++;
if(no - st == -1)s[no]++;
}
else if(no - st == 0)
{
a[0][1]++;
a[1][1]++;
}
else
{
a[0][2]++;
a[1][0]++;
if(no - st == 7)s1[st]++;
if(no - st == -8)s1[st]++;
if(no - st == 1)s1[st]++;
}
}
/*cout << s[66] << s[67] << s[74] << endl;
cout << s1[66] << s1[67] <<s1[74] << endl;*/
if(max < s[66]||max1 < s1[66]){
if(max1 < s1[66]){max1=s1[66];b=66;}
if(max < s[66]){max=s[66];a1=66;}}
if(max < s[67]||max1 < s1[67]){
if(max < s[67] ){max=s[67];t=1;a1=67;}
if(max1 < s1[67]){max1=s1[67];t1=1;b=67;}}
if((max < s[74]||max1 < s1[74])){
if(t != 1 && max < s[74]){max=s[74];a1=74;}
if(t1 != 1 && max1 < s1[74]){max1=s1[74];b=74;}}
for(i = 0; i < 2; i++)
{
for(j = 0; j < 3; j++)
{if(j != 2)cout << a[i][j] <<' ';
else cout << a[i][j];
}
cout << endl;
}
if(a1==66)cout << 'B' << ' ';
if(a1==67)cout << 'C' << ' ';
if(a1==74)cout << 'J' << ' ';
if(b==66)cout << 'B';
if(b==67)cout << 'C';
if(b==74)cout << 'J';
return 0;
}
using namespace std;
int main()
{
int m,i,j,s[80]={0},s1[80]={0},t=0,t1=0,t2=0,a[2][3]={0},max=0,max1=0,a1,b;
char no,st;
cin >> m;
for(i = 0; i < m; i++)
{
cin >> no >>st;
if(no - st == -7 || no - st == 8 || no - st == -1)
{
a[0][0]++;
a[1][2]++;
if(no - st == -7)s[no]++;
if(no - st == 8)s[no]++;
if(no - st == -1)s[no]++;
}
else if(no - st == 0)
{
a[0][1]++;
a[1][1]++;
}
else
{
a[0][2]++;
a[1][0]++;
if(no - st == 7)s1[st]++;
if(no - st == -8)s1[st]++;
if(no - st == 1)s1[st]++;
}
}
/*cout << s[66] << s[67] << s[74] << endl;
cout << s1[66] << s1[67] <<s1[74] << endl;*/
if(max < s[66]||max1 < s1[66]){
if(max1 < s1[66]){max1=s1[66];b=66;}
if(max < s[66]){max=s[66];a1=66;}}
if(max < s[67]||max1 < s1[67]){
if(max < s[67] ){max=s[67];t=1;a1=67;}
if(max1 < s1[67]){max1=s1[67];t1=1;b=67;}}
if((max < s[74]||max1 < s1[74])){
if(t != 1 && max < s[74]){max=s[74];a1=74;}
if(t1 != 1 && max1 < s1[74]){max1=s1[74];b=74;}}
for(i = 0; i < 2; i++)
{
for(j = 0; j < 3; j++)
{if(j != 2)cout << a[i][j] <<' ';
else cout << a[i][j];
}
cout << endl;
}
if(a1==66)cout << 'B' << ' ';
if(a1==67)cout << 'C' << ' ';
if(a1==74)cout << 'J' << ' ';
if(b==66)cout << 'B';
if(b==67)cout << 'C';
if(b==74)cout << 'J';
return 0;
}
编辑于 2018-08-14 09:44:30
回复(0)
//// p_1008.cpp// PAT_basic_level//// Created by 李嘉宁 on 2018/4/19.// Copyright © 2018年 李嘉宁. All rights reserved.//#include <iostream>using namespace std;intmain(){ints = 0, p = 0, f = 0;intc1 = 0, j1 = 0, b1 = 0;intc2 = 0, j2 = 0, b2 = 0;intN = 0;charjia[105], yi[105]; // 考虑过二维数组cin >> N;for(inti = 0; i < N; i++){cin >> jia[i] >> yi[i];if(jia[i] == 'C'&& yi[i] == 'J') // 考虑过C\J\B 之间的规律{s++;c1++;}elseif(jia[i] == 'J'&& yi[i] == 'B'){s++;j1++;}elseif(jia[i] == 'B'&& yi[i] == 'C'){s++;b1++;}elseif((jia[i] == 'C'&& yi[i] == 'C') || (jia[i] == 'J'&& yi[i] == 'J') || (jia[i] == 'B'&& yi[i] == 'B')){p++;}elseif(jia[i] == 'C'&& yi[i] == 'B'){f++;b2++;}elseif(jia[i] == 'J'&& yi[i] == 'C'){f++;c2++;}elseif(jia[i] == 'B'&& yi[i] == 'J'){f++;j2++;}}cout << s << ' '<< p << ' '<< f << endl;cout << f << ' '<< p << ' '<< s << endl;cout << ((((b1 >= c1) ? b1 : c1) >= j1) ? ((b1 >= c1) ? 'B': 'C') : 'J') << ' '<< ((((b2 >= c2) ? b2 : c2) >= j2) ? ((b2 >= c2) ? 'B': 'C') : 'J');return0;}
发表于 2018-04-19 15:24:28
回复(0)
j_p = 0
j_s_c, j_s_j, j_s_b = 0, 0, 0
y_p = 0
y_s_c, y_s_j, y_s_b = 0, 0, 0
def most_win(b, c, j):
if b == max(b, c, j):
return 'B'
elif c == max(b, c, j):
return 'C'
elif j == max(b, c, j):
return 'J'
n = int(input())
for i in range(n):
j, y = input().split()
if j == y:
j_p += 1 y_p += 1
elif j == 'C' and y == 'J':
j_s_c += 1
elif j == 'J' and y == 'B':
j_s_j += 1
elif j == 'B' and y == 'C':
j_s_b += 1
elif y == 'B' and j == 'C':
y_s_b += 1
elif y == 'C' and j == 'J':
y_s_c += 1
elif y == 'J' and j == 'B':
y_s_j += 1
j_s = j_s_b + j_s_c + j_s_j
j_f = n - j_p - j_s
y_s = y_s_b + y_s_c + y_s_j
y_f = n - y_p - y_s
j_s_most = most_win(j_s_b, j_s_c, j_s_j)
y_s_most = most_win(y_s_b, y_s_c, y_s_j)
print(j_s, j_p, j_f)
print(y_s, y_p, y_f)
print(j_s_most, y_s_most)
编辑于 2017-11-14 14:57:20
回复(0)
#include <iostream>
using namespace std; int compare(char a,char b)
{
if((a=='B'&&b=='C')||(a=='C'&&b=='J')||(a=='J'&&b=='B'))
return 1;//a获胜
if((a=='B'&&b=='B')||(a=='C'&&b=='C')||(a=='J'&&b=='J'))
return 0;//平
if((a=='C'&&b=='B')||(a=='J'&&b=='C')||(a=='B'&&b=='J'))
return -1;//b获胜
return 0;
} int judge(char a)
{
if(a=='B')
return 1;
if(a=='C')
return 2;
if(a=='J')
return 3;
return 0;
} char max1(int b,int c,int j)
{
if(b!=c&&b!=j&&c!=j){
if(b>c&&b>j)
return 'B';
if(c>b&&c>j)
return 'C';
if(j>b&&j>c)
return 'J';
}else{
if(b==c&&b>j)
return 'B';
if(b==c&&b<j)
return 'J';
if(c==j&&c>b)
return 'C';
if(c==j&&c<b)
return 'B';
if(b==j&&b>c)
return 'B';
if(b==j&&b<c)
return 'C';
if(b==j&&c==j&&b==c)
return 'B';
}
return 'A';
} int main()
{
int count1,count2,count3,n,i,c1,j1,b1,b2,c2,j2;
count1=count2=count3=0;
c1=j1=b1=b2=c2=j2=0;
char a,b;
cin >> n;
for(i=0;i<n;i++){
cin >> a >> b;
if(compare(a,b)==1){
count1++;
if(judge(a)==1)
b1++;
if(judge(a)==2)
c1++;
if(judge(a)==3)
j1++;
}
if(compare(a,b)==0)
count2++;
if(compare(a,b)==-1){
count3++;
if(judge(b)==1)
b2++;
if(judge(b)==2)
c2++;
if(judge(b)==3)
j2++;
}
}
//cout << b1 << c1 << j1 <<endl;
// cout << b2 << c2 << j2 <<endl;
cout << count1 << ' ' << count2 << ' ' << count3 << endl;
cout << count3 << ' ' << count2 << ' ' << count1 << endl;
cout << max1(b1,c1,j1) << ' ' << max1(b2,c2,j2) << endl;
return 0;
}
发表于 2017-08-09 12:06:30
回复(0)
#encoding:utf-8
#C:锤子，J:剪刀，B：布
# c>j,j>b,b>c;a1,a2列表作为优势列表;r1,r2作为胜负结果列表(胜，平，负)
a1 = {'J':0,'C':0,'B':0}
a2 = {'J':0,'C':0,'B':0}
r1 = [0,0,0]
r2 = [0,0,0]
tmp = ''
y1=y2=''
times = int(raw_input())
if times:
for v in range(times):
t1,t2 = raw_input().split(' ')
if t1 == 'C' and t2 == 'J':
r1[0]+=1
r2[2]+=1
a1['C']+=1
# a2['J']+=1
elif t1 == 'J' and t2 =='C':
r2[0]+=1
r1[2]+=1
a2['C']+=1
# a1['J']+=1
elif t1 == 'J' and t2 == 'B':
r1[0]+=1
r2[2]+=1
a1['J']+=1
elif t1 =='B' and t2 == 'J':
r2[0]+=1
r1[2]+=1
a2['J'] +=1
elif t1 == 'B' and t2 =='C':
r1[0]+=1
r2[2]+=1
a1['B']+=1
elif t1 == 'C' and t2 == 'B':
r2[0]+=1
r1[2]+=1
a2['B']+=1
else:
r1[1]+=1
r2[1]+=1
a1 = sorted(a1.iteritems(),key=lambda x:x,reverse=True)
a2 = sorted(a2.iteritems(),key=lambda x:x,reverse=True)
for k,v in a1:
if v == 0:
continue
v = tmp
y1 = k
if v>tmp:
c = tmp
tmp = v
v = c
y1 = k
for k,v in a2:
if v == 0:
continue
v = tmp
y2 = k
if v>tmp:
c = tmp
tmp = v
v = c
y2 = k
if not y1:
y1 = 'B'
if not y2:
y2 = 'B'
print '%d %d %d' % tuple(r1)
print '%d %d %d' % tuple(r2)
print '%s %s' % (y1,y2)
发表于 2017-04-11 23:00:37
回复(0)
import java.util.Scanner;
public class Main {
/*
* 数组存储，做差求值，下标输出
*/
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] num = new int[3];// num[0]:甲胜 num[1]:平局 num[2]:乙胜
char[] b = { 'B', 'C', 'J' };// 哈哈，正好为字母表顺序存储
int[][] bt = new int[3][2];// bt[i][0]:甲BCJ胜的次数 乙同理
char[] w = new char[2];
for (int i = 0; i < n; i++) {
char t1 = in.next().charAt(0);
char t2 = in.next().charAt(0);
int te1 = -10, te2 = -10;
for (int j = 0; j < b.length; j++) {
if (t1 == b[j])
te1 = j;
if (t2 == b[j])
te2 = j;
}
switch (te1 - te2) {
case -1:// 甲胜
case 2:
num[0]++; bt[te1][0]++; break;
case 1:// 乙胜
case -2:
num[2]++; bt[te2][1]++; break;
default:// 平局
num[1]++; break;
}
}
System.out.printf("%d %d %d\n", num[0], num[1], num[2]);
System.out.printf("%d %d %d\n", num[2], num[1], num[0]);
for (int j = 0; j < 2; j++) {
int max = bt[0][j];
int ww = 0;
for (int i = 1; i < 3; i++) {
if (bt[i][j] > max) {
max = bt[i][j];
ww = i;
}
}
System.out.print(b[ww]);
if (j != 1)
System.out.print(" ");
}
}
}
发表于 2017-01-15 18:46:02
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#include<iostream>using namespace std;voidfight(charplayerA,charplayerB);intidexMax(intWinA[3]);intplayResult[3]={0,0,0},winA[3]={0,0,0},winB[3]={0,0,0};intmain(){intN;charplayerA,playerB;charindex[3]={'C','J','B'};cin>>N;while(N--){cin>>playerA>>playerB;fight(playerA,playerB);}cout<<playResult[0]<<" "<<playResult[1]<<" "<<playResult[2]<<endl;cout<<playResult[2]<<" "<<playResult[1]<<" "<<playResult[0]<<endl;cout<<index[idexMax(winA)]<<" "<<index[idexMax(winB)]<<endl;return0;}//save the player fight result.voidfight(charplayerA,charplayerB){if(playerA==playerB)playResult[1]++;else{switch(playerA){case'C':if(playerB=='B'){//B uses the 'B' win the playerplayResult[2]++;winB[2]++;}else{playResult[0]++;//A uses the 'C' win the playerwinA[0]++;}break;case'J':if(playerB=='C'){//B uses the 'C' win the playerplayResult[2]++;winB[0]++;}else{playResult[0]++;//A uses the 'J' win the playerwinA[1]++;}break;case'B':if(playerB=='J'){//B uses the 'J' win the playerplayResult[2]++;winB[1]++;}else{playResult[0]++;//A uses the 'C' win the playerwinA[2]++;}break;}}}intidexMax(intwin[3]){inttmp=2;if((win[2]>=win[0])&&(win[2]>=win[1])){tmp = 2;//output 'B' by priorityreturn2;}elseif((win[0]>=win[1])&&(win[0]>=win[2])){tmp = 0;//output 'C'return0;}elseif((win[1]>=win[0])&&(win[1]>=win[2])){tmp = 1;//output 'J'return1;}returntmp;}
发表于 2017-01-15 17:01:10
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#include<stdio.h>
int main()
{
char a,b,Most[3]={'B','C','J'};
int N,i,compare;
int RA[3]={0},A[3]={0},B[3]={0};
scanf("%d\n",&N);
for(i=0;i<N;i++){
scanf("%c %c",&a,&b);
getchar();
compare=(int)(a-b);
if(compare==0)
RA[1]++;
if(compare==-7){
RA[0]++;
A[1]++;
}
if(compare==7){
RA[2]++;
B[1]++;
}
if(compare==1){
RA[2]++;
B[0]++;
}
if(compare==-1){
RA[0]++;
A[0]++;
}
if(compare==8){
RA[0]++;
A[2]++;
}
if(compare==-8){
RA[2]++;
B[2]++;
}
}
printf("%d %d %d\n",RA[0],RA[1],RA[2]);
printf("%d %d %d\n",RA[2],RA[1],RA[0]);
N=0;
compare=0;
for(i=1;i<3;i++){
if(A[i]>A[N])
N=i;
if(B[i]>B[compare])
compare=i;
}
printf("%c ",Most[N]);
printf("%c",Most[compare]);
return 0;
}
发表于 2017-01-14 14:04:44
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