[编程题]打牌
- 热度指数：12999 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
- 算法知识视频讲解
牌只有1到9，手里拿着已经排好序的牌a，对方出牌b，用程序判断手中牌是否能够压过对方出牌。
规则：出牌牌型有5种
[1]一张 如4 则5...9可压过
[2]两张 如44 则55，66，77，...，99可压过
[3]三张 如444 规则如[2]
[4]四张 如4444 规则如[2]
[5]五张 牌型只有12345 23456 34567 45678 56789五个，后面的比前面的均大。
输入描述:
输入有多组数据。 每组输入两个字符串(字符串大小不超过100)a，b。a字符串代表手中牌，b字符串代表出的牌。
输出描述:
压过输出YES 否则NO。
示例1
输入
12233445566677 33
输出
YES
#include <cstdio>
#include <cstring>
#include <memory>
const int N = 100;
int main()
{
char a[N], b[6];
int num[10];
while (scanf("%s%s", a, b) != EOF)
{
memset(num, 0, sizeof(num));
int len_a = strlen(a);
int len_b = strlen(b);
bool flag = false;
for (int i = 0; i < len_a; i++)
{
num[a[i] - '0']++;
}
if (len_b < 5)
{
for (int i = b[0] - '0' + 1; i < 10; i++)
{
if (num[i] >= len_b)
{
flag = true;
break;
}
}
}
else
{
for (int i = b[0] - '0' + 1; i < 6; i++)
{
if (num[i] > 0 && num[i + 1] > 0 && num[i + 2] > 0
&& num[i + 3] > 0 && num[i + 4] > 0)
{
flag = true;
break;
}
}
}
if (flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
发表于 2017-06-06 14:05:59
回复(1)
# include<stdio.h>
# include<string.h>
int main()
{
//手中牌
char str[101];
//待处理的牌
char card[10];
//计数数组，代表每个数字有多少个，‘0’记录在arr[0]中
int arr[10];
while(scanf("%s",str)!=EOF)
{
scanf("%s",card);
int s_len=strlen(str);
int c_len=strlen(card);
for(int i=0;i<10;i++)
{
arr[i]=0;
}
//'0'的ascii码表的十进制是48
for(int i=0;i<s_len;i++)
{
arr[str[i]-48]++;
}
//情况1
if(c_len==1)
{
//flag==0为NO
int flag=0;
for(int i=0;i<10;i++)
{
if(arr[i]>0&&i>card[0]-48)
{
flag=1;
break;
}
}
if(flag==1) printf("YES\n");
else printf("NO\n");
}
//情况2
if(c_len==2)
{
//flag==0为NO
int flag=0;
for(int i=0;i<10;i++)
{
if(arr[i]>1&&i>card[0]-48)
{
flag=1;
break;
}
}
if(flag==1) printf("YES\n");
else printf("NO\n");
}
//情况3
if(c_len==3)
{
//flag==0为NO
int flag=0;
for(int i=1;i<10;i++)
{
if(arr[i]>2&&i>card[0]-48)
{
flag=1;
//printf("%d %d\n",arr[i],i);
break;
}
}
if(flag==1) printf("YES\n");
else printf("NO\n");
}
//情况4
if(c_len==4)
{
//flag==0为NO
int flag=0;
for(int i=0;i<10;i++)
{
if(arr[i]>3&&i>card[0]-48)
{
flag=1;
break;
}
}
if(flag==1) printf("YES\n");
else printf("NO\n");
}
//情况5
if(c_len==5)
{
//flag==0为NO
int flag=0;
for(int i=2;i<=5;i++)
{
//连续5个数字，并且要求数字数量大于1
if(arr[i]>0&&arr[i+1]>0&&arr[i+2]>0&&arr[i+3]>0&&arr[i+4]>0&&
i>card[0]-48)
{
flag=1;
break;
}
}
if(flag==1) printf("YES\n");
else printf("NO\n");
}
}
return 0;
}
发表于 2018-07-17 01:20:46
回复(1)
#include<iostream>
#include<string>
using namespace std;
int main()
{
int n, i, count[10];
string a, b;
bool flag = false;
while(cin >> a >> b)
{
for(i = 0;i < 10;++i)
count[i] = 0;
for(i = 0;i < a.size();++i) //统计数量
++count[a[i] - '0'];
if(b.size() < 5) //前四个一种情况，看数量
{
for(i = b[0] + 1 -'0';i < 10;++i)
{
if(count[i] >= b.size())
{
flag = true;
break;
}
}
}
else //顺子，看连续
{
n = 0;
for(i = b[1] - '0';i < 10;++i)
{
if(count[i] != 0)
++n;
else
n = 0;
if(n == 5)
{
flag = true;
break;
}
}
}
if(flag)
cout << "YES" << endl;
else
cout << "NO" <<endl;
}//while
return 0;
}
编辑于 2019-03-03 04:37:43
回复(0)
#include<stdio.h> #include<string.h> //将每张牌的数量进行统计，分为2中牌型进行求解 int case1(int a[],int n,int length){ int i; for(i=8;i>=0;i--){ //i+1为牌的牌面 if(n<i+1&&length<=a[i]) return 1; } return 0; } int case2(int a[],int n){ int i; for(i=8;i>=4;i--){ if(i+1>n&&a[i]>0&&a[i-1]>0&&a[i-2]>0&&a[i-3]>0&&a[i-4]>0) return 1; } return 0; }int main(){ char b[100],c[100]; int result,a[9]={0},i,lengthb,lengthc; while(scanf("%s%s",b,c)!=EOF){ lengthb=strlen(b); lengthc=strlen(c); for(i=0;i<lengthb;i++) a[b[i]-48-1]++; if(lengthc<=4) result=case1(a,c[0]-48,lengthc); else result=case2(a,c[4]-48); if(result==1) printf("YES\n"); else printf("NO\n"); } }
编辑于 2019-02-19 21:49:08
回复(2)
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
//数字转字符串 常用
string itoa(int i){
stringstream s;
s << i;
return s.str();
}
//***1、2、3、4种情况
bool matching(string a, int _b, int type) {
string _str;
for (int i = _b + 1; i <= 9; ++i){
_str = "";
for (int j = 0; j < type; ++j){
_str += itoa(i);
}
if (a.find(_str) != a.npos)
return true;
}
return false;
}
int main(){
string a, b;
cin >> a >> b;
int type = b.size();
bool result = false;
//判断是哪种类型的出牌
int _b = b[0] - '0'; //char 转成 int
//单独***
if (type == 5){
string _str;
bool flag = false;
if (_b <= 5){
for (int i = _b + 1; i <= 5; ++i){
int _c = i;
int j = 0;
for (; j < 5; ++j, ++_c){
if (a.find(itoa(_c)) == a.npos){ //假如有一个数字没找到就直接跳出这一循环
break;
}
}
if (j == 5) //表示联系的5个数字都找到
flag = true;
}
//标志为真 则查找成功
if (flag)
result = true;
}
}
else {
result = matching(a, _b, type);
}
if (result)
cout << "YES";
else
cout << "NO";
return 0;
}
发表于 2018-02-07 18:07:41
回复(0)
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main(void)
{
string a,b;
while(cin >> a >> b)
{
sort(a.begin(),a.end());
while(1)//这个循环是为了防止b + 1后在a中没找到，继续加1直到找到能压过b的字符串
{
if(b[0] == '9')
{
cout << "NO" << endl;//当为9的时候牌已经最大了，压不过
break;
}
for(int i = 0;i < b.size();i++)//更新比b大的字符串
b[i] += 1;
//find(b)函数用于寻找b字符串并返回对应字符串的首字符的数组下标，没找到则返回一个npos
if(a.find(b) != a.npos)//对子或叠牌的情况
{
cout << "YES" << endl;
break;
}
string c = a;
unique(c.begin(), c.end());//去掉重复的相邻字符,前提是已经排序好的字符串
if(c.find(b) != c.npos)//顺子的情况如:12345
{
cout << "YES" << endl;
break;
}
}
}
return 0;
}
发表于 2021-02-16 14:34:50
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()) {
String str1 = scanner.next();
String str2= scanner.next();
int[] nums = new int[str2.length()];
for(int i = 0; i < str2.length(); i++) {
nums[i] = ((int)str2.charAt(i) - 48);
}
if(nums.length == 5){
int i = 0, j = 1;
while(i < 5) {
while(!(str1.contains(String.valueOf(nums[i] + j))) && j != 5) {
i = 0;
j++;
}
if(j == 5) {
System.out.println("NO");
break;
}
if(i < 4) i++;
else {
System.out.println("YES");
break;
}
}
}else {
int i = 0, j = 0, sum = Integer.parseInt(str2);
if(str2.length() == 4) i = 1111;
if(str2.length() == 3) i = 111;
if(str2.length() == 2) i = 11;
if(str2.length() == 1) i = 1;
for(j = 0; j < 8; j++) {
sum += i;
if(str1.contains(String.valueOf(sum))) {
System.out.println("YES");
break;
}
}
if(j == 8)System.out.println("NO");
}
}
}
}
public class Main{
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()) {
String str1 = scanner.next();
String str2= scanner.next();
int[] nums = new int[str2.length()];
for(int i = 0; i < str2.length(); i++) {
nums[i] = ((int)str2.charAt(i) - 48);
}
if(nums.length == 5){
int i = 0, j = 1;
while(i < 5) {
while(!(str1.contains(String.valueOf(nums[i] + j))) && j != 5) {
i = 0;
j++;
}
if(j == 5) {
System.out.println("NO");
break;
}
if(i < 4) i++;
else {
System.out.println("YES");
break;
}
}
}else {
int i = 0, j = 0, sum = Integer.parseInt(str2);
if(str2.length() == 4) i = 1111;
if(str2.length() == 3) i = 111;
if(str2.length() == 2) i = 11;
if(str2.length() == 1) i = 1;
for(j = 0; j < 8; j++) {
sum += i;
if(str1.contains(String.valueOf(sum))) {
System.out.println("YES");
break;
}
}
if(j == 8)System.out.println("NO");
}
}
}
}
发表于 2020-03-07 19:05:11
回复(0)
#include<stdio.h>
#include<string.h>
int main(){
int a[20];
memset(a,0,sizeof(a));
char ans[100];
gets(ans);
int le=strlen(ans);
for(int i=0;i<le;i++)
a[ans[i]-48]++;//经大神指点，字符串和数字以‘0’-48=0转换
char b[6];
gets(b);
int l=strlen(b);
int i=b[0]-48+1;
switch(l){
case 1:
if(b[0]-48==9)
printf("NO");
else{
for(i=b[0]-48+1;i<10;i++){
if(a[i]){
printf("YES");
break;
}
}
if(i==10)
printf("NO");
}
break;
case 2:
if(b[0]-48==9)
printf("NO");
else{
for(i=b[0]-48+1;i<10;i++){
if(a[i]>1){
printf("YES");
break;
}
}
if(i==10)
printf("NO");
}
break;
case 3:
if(b[0]-48==9)
printf("NO");
else{
for(i=b[0]-48+1;i<10;i++){
if(a[i]>2){
printf("YES");
break;
}
}
if(i==10)
printf("NO");
}
break;
case 4:
if(b[0]-48==9)
printf("NO");
else{
for(i=b[0]-48+1;i<10;i++){
if(a[i]>3){
printf("YES");
break;
}
}
if(i==10)
printf("NO");
}
break;
case 5:
if(b[0]-48==5)
printf("NO");
else{
for(i=b[0]-48+1;i<6;i++){
if(a[i]&&a[i+1]&&a[i+2]&&a[i+3]&&a[i+4]){
printf("YES");
break;
}
}
if(i==6)
printf("NO");
}
break;
}
return 0;
}
发表于 2019-01-23 18:26:40
回复(0)
#include<bits/stdc++.h>
#define ll long long
#define len 5
using namespace std;
string a,b;
int change[105];
int main()
{
std::ios::sync_with_stdio(false);
while(cin>>a>>b)
{
for(int i=0;i<a.size();i++)
{
change[a[i]-'0']++;
}
if(b.size()<5)
{
for(int i=0;i<a.size();i++)
{
if(a[i]>b[0]&&change[a[i]-'0']>=b.size())
{
cout<<"YES"<<endl;
break;
}
if(i==a.size()-1)
cout<<"NO"<<endl;
}
}
else
{
for(int j=0,i;j<a.size()-4;j++)
{
i=a[j]-'0';
if(a[j]>b[0]&&change[i]>0&&change[i+1]>0&&change[i+2]>0&&change[i+3]>0&&change[i+4]>0)
{
cout<<"YES"<<endl;
break;
}
if(j==a.size()-5)
cout<<"NO"<<endl;
}
}
memset(change,0,sizeof(change));
}
return 0;
}
发表于 2018-05-24 12:51:25
回复(0)
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int a[10]= {0};
int main()
{
string str,s;
while( cin>>str>>s)
{
for(int i=0; i<str.size(); i++)
{
int num=str[i]-48;
a[num]++;
}
if(s.size()<=4)
{
int num=s[0]-48;
int i;
for( i=num+1; i<=9; i++)
{
if(a[i]>=s.size())
{
cout<<"YES"<<endl;
break;
}
}
if(i>9) cout<<"NO"<<endl;
}
else if(s.size()==5)
{
int num=s[0]-48;
int flag=0,i;
for( i=num+1; i<=9; i++)
{
if(a[i]>=1)
{
flag++;
if(flag>=5)
{
cout<<"YES"<<endl;
break;
}
}
else
flag=0;
}
if(i>9) cout<<"NO"<<endl;
}
}
}
发表于 2018-03-24 21:22:52
回复(0)
利用hash数组标记再比较
#include <iostream>
#include<string.h>using namespace std;
int main()
{
char a[100],b[5];
while(cin>>a>>b){
int hash[10]={0};
int x=strlen(a);
int y=strlen(b);
for(int i=0;i<x;i++)
hash[(a[i]-'0')]++;
int count=0;
int flag=b[0]-'0';
if(y<5){
for(int i=flag+1;i<10;i++){
if(hash[i]>=y)
{
count=1;
break;
}
}
}
else{
for(int i=flag;i+5<10;i++){
if(hash[i+1]>=1&&hash[i+2]>=1&&hash[i+3]>=1&&hash[i+4]>=1&&hash[i+5]>=1)
count=1;
}
}
if(count==1)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
发表于 2018-03-24 13:57:52
回复(0)
#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{ char a[100],b[100],c[100],d[100];
for(int i=0;i<100;i++){
a[i]=0;
b[i]=0;
c[100]=0;
d[100]=0;
}//初始化
scanf("%s",a);
scanf("%s",b);
int numa=0,numb=0;
for(int i=0;a[i]!='\0';i++){
if(a[i]>='1'&&a[i]<='9')
numa++;
}
for(int i=0;b[i]!='\0';i++){
if(b[i]>='1'&&b[i]<='9')
numb++;
} //字母个数正确
if(numb==1){
if(a[numa-1]>b[0])//设计精髓，最后一个大于即可以
printf("YES\n");
else printf("NO\n");
}
else if(numb==2){
int number1=0;
for(int i=1;i<numa;i++){
if(a[i]==a[i-1])
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==3){
int number1=0;
for(int i=1;i<numa;i++){
if((a[i]==a[i-1])&&(a[i]==a[i-2]))
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==4){
int number1=0;
for(int i=1;i<numa;i++){
if((a[i]==a[i-1])&&(a[i]==a[i-2])&&(a[i]==a[i-3]))
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==5){
int number1=0,number2=0;
char e=0;
for(int i=0;i<numa;i++){
if(a[i]!=a[i+1]){
if(a[i]>e){
d[number2++]=a[i];
e=a[i];
}
}
}//去掉a中重复字符
for(int i=0;i<number2;i++){
if((d[i]==d[i+1]-1)&&(d[i]==d[i+2]-2)&&(d[i]==d[i+3]-3)&&(d[i]==d[i+4]-4))
c[number1++]=d[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
system("PAUSE");
return EXIT_SUCCESS;
}
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{ char a[100],b[100],c[100],d[100];
for(int i=0;i<100;i++){
a[i]=0;
b[i]=0;
c[100]=0;
d[100]=0;
}//初始化
scanf("%s",a);
scanf("%s",b);
int numa=0,numb=0;
for(int i=0;a[i]!='\0';i++){
if(a[i]>='1'&&a[i]<='9')
numa++;
}
for(int i=0;b[i]!='\0';i++){
if(b[i]>='1'&&b[i]<='9')
numb++;
} //字母个数正确
if(numb==1){
if(a[numa-1]>b[0])//设计精髓，最后一个大于即可以
printf("YES\n");
else printf("NO\n");
}
else if(numb==2){
int number1=0;
for(int i=1;i<numa;i++){
if(a[i]==a[i-1])
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==3){
int number1=0;
for(int i=1;i<numa;i++){
if((a[i]==a[i-1])&&(a[i]==a[i-2]))
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==4){
int number1=0;
for(int i=1;i<numa;i++){
if((a[i]==a[i-1])&&(a[i]==a[i-2])&&(a[i]==a[i-3]))
c[number1++]=a[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
else if(numb==5){
int number1=0,number2=0;
char e=0;
for(int i=0;i<numa;i++){
if(a[i]!=a[i+1]){
if(a[i]>e){
d[number2++]=a[i];
e=a[i];
}
}
}//去掉a中重复字符
for(int i=0;i<number2;i++){
if((d[i]==d[i+1]-1)&&(d[i]==d[i+2]-2)&&(d[i]==d[i+3]-3)&&(d[i]==d[i+4]-4))
c[number1++]=d[i];
}
if(c[number1-1]>b[0])
printf("YES");
else printf("NO");
}
system("PAUSE");
return EXIT_SUCCESS;
}
发表于 2018-03-13 15:17:43
回复(0)
编到一半，发现哈希计数很简单，不过，我还是把我第一次的思路实现了。 swicth的重复可以提取出来，自行优化
package com.speical.first;
import java.util.Scanner;
public class Pro190 {
public static boolean compare(String strA, String strB) {
boolean flag = false;
int index = 0;
for(; index < strA.length(); index++) {
if(strA.charAt(index) > strB.charAt(0)) {
break;
}
}
int count;
if(index < strA.length()) {
switch (strB.length()) {
case 1: flag = true; break;
case 2:
for(int j = index; j < strA.length();) {
if((++j) < strA.length() && strA.charAt(j) == strA.charAt(j - 1)) {
flag = true;
break;
}
}
break;
case 3:
for(int j = index; j < strA.length();) {
count = 2;
while((++j) < strA.length() && strA.charAt(j) == strA.charAt(j - 1)) {
count--;
}
if(count == 0) {
flag = true;
break;
}
}
break;
case 4:
for(int j = index; j < strA.length();) {
count = 3;
while((++j) < strA.length() && strA.charAt(j) == strA.charAt(j - 1)) {
count--;
}
if(count == 0) {
flag = true;
break;
}
}
break;
case 5:
char ch;
for(int j = index; j < strA.length(); j++) {
count = 4;
ch = strA.charAt(j);
while(strA.indexOf((char)(++ch)) != -1 && count-- > 0);
if(count == 0) {
flag = true;
break;
}
}
break;
}
}
return flag;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
while(input.hasNext()) {
String strA = input.nextLine();
String strB = input.nextLine();
System.out.println(compare(strA, strB) ? "YES" : "NO");
}
}
}
发表于 2018-02-04 22:07:46
回复(0)
#include <bits/stdc++.h>
using namespace std;
int a[10];
int main()
{
string s,s1;
while(cin>>s>>s1)
{
for(int i=0;i<s.size();i++)
{
a[s[i]-'0']++;
}
//五张牌特殊处理
if(s1.size()==5)
{
int tag=0;
for(int i=s1[0]-'0'+1;i<=5;i++)//以i开头连续5个
{
int flag=1;
for(int j=0;j<=4;j++)
{
if(a[i+j]==0)
{
flag=0;//中间某张没有
break;
}
}
if(flag) tag=1;//五张都有
}
if(tag)
{
cout<<"YES"<<endl;
}
else cout<<"NO"<<endl;
}
else
{
int flag=0;
for(int j=s1[0]-'0'+1;j<=9;j++)
{
if(a[j]>=s1.size())
{
flag=1;
break;
}
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}
编辑于 2024-03-21 17:28:35
回复(0)
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main() {
string a, b;
while (cin >> a >> b) {
vector<int>number(10); //各个数字在字符串a中的个数
for (const auto& ch : a) {
number[ch - '0']++;
}
int n = b.length();
bool flag = false;
if (n <= 4) {
for (int i = b[0] - '0' + 1; i <= 9; i++) {
if (number[i] >= n) {
flag = true;
break;
}
}
} else {
int count = 0;
if (b[0] >= '5') {
flag = false;
} else {
for (int i = b[0] - '0' + 1; i <= 9 && count < 5; i++) {
if (number[i] > 0) {
count++;
} else {
count = 0;
}
}
flag = count == 5;
}
}
cout << (flag ? "YES" : "NO") << endl;
}
return 0;
}
编辑于 2024-03-07 19:10:59
回复(0)
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String []args) {
Scanner in = new Scanner(System.in);
while (in.hasNextLine()) {
String a = in.nextLine(); //表示手中的牌
String b = in.nextLine(); //表示出的牌
boolean flag = false; //标识 是否有大过的牌
if (b.length() == 1) { //只有一张牌
char c = b.charAt(0); //读取第一个字符
//判断a中是否有大过c的牌
for (int i = 0; i < a.length(); i++) {
if (a.charAt(i) > c) {
flag = true;
break;
}
}
if (flag) { //能压下去
System.out.println("YES");
flag = false; //重置标识
continue;
} else {
System.out.println("NO");
}
} else if (b.length() >= 2 && b.length() <= 4) { //"对子"、"三连"、"炸"
char c = b.charAt(0);
for (int i = 0; i < a.length(); i++) { //查看是否有大过的牌
if (a.charAt(i) > c) {
int length = getCharacterLength(a, a.charAt(i)); //该字母的牌数
if (length == 1) { //牌只有一张，不满足条件
continue;
}
if (b.length() == 2) {
if (length >= 2) {
flag = true;
break;
}
} else if (b.length() == 3) {
if (length >= 3) {
flag = true;
break;
}
} else {
if (length == 4) {
flag = true;
break;
}
}
}
}
if (flag) {
flag = false; //重置标识
System.out.println("YES");
continue;
} else {
System.out.println("NO");
}
} else { //五张"顺子"
int all_character[] = new int[10];
for (int i = 0; i < 10; i++)
all_character[i] = 0;
for (int i = 0; i < a.length(); i++) {
all_character[Integer.parseInt(Character.toString(a.charAt(
i)))]++; //对应数字 个数+1
}
int record = Integer.parseInt(Character.toString(b.charAt(0)));
if (b.charAt(0) == '1') { //为123456
if (all_character[record + 1] != 0 && all_character[record + 2] != 0 &&
all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0
|| all_character[record + 2] != 0 && all_character[record + 3] != 0 &&
all_character[record + 4] != 0 && all_character[record + 5] != 0 &&
all_character[record + 6] != 0
|| all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0 && all_character[record + 6] != 0 &&
all_character[record + 7] != 0
|| all_character[record + 4] != 0 && all_character[record + 5] != 0 &&
all_character[record + 6] != 0 && all_character[record + 7] != 0 &&
all_character[record + 8] != 0) {
flag = true;
}
} else if (b.charAt(0) == '2') {
if (all_character[record + 1] != 0 && all_character[record + 2] != 0 &&
all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0
|| all_character[record + 2] != 0 && all_character[record + 3] != 0 &&
all_character[record + 4] != 0 && all_character[record + 5] != 0 &&
all_character[record + 6] != 0
|| all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0 && all_character[record + 6] != 0 &&
all_character[record + 7] != 0) {
flag = true;
}
} else if (b.charAt(0) == '3') {
if (all_character[record + 1] != 0 && all_character[record + 2] != 0 &&
all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0
|| all_character[record + 2] != 0 && all_character[record + 3] != 0 &&
all_character[record + 4] != 0 && all_character[record + 5] != 0 &&
all_character[record + 6] != 0) {
flag = true;
}
} else if (all_character[record + 1] != 0 && all_character[record + 2] != 0 &&
all_character[record + 3] != 0 && all_character[record + 4] != 0 &&
all_character[record + 5] != 0) {
if (a.contains("56789")) {
flag = true;
}
}
if (flag) {
flag = false; //重置标识
System.out.println("YES");
continue;
} else {
System.out.println("NO");
}
}
}
}
/**
* 得到某个字符总长度
* @param a
* @param c
* @return
*/
public static int getCharacterLength(String a, char c) {
int length = 0;
for (int i = 0; i < a.length(); i++) {
if (a.charAt(i) == c)
length++;
}
return length;
}
}
发表于 2023-03-10 09:40:10
回复(0)
我最短！
#include <iostream>
#include <string>
#include <array>
#include <algorithm>
using namespace std;
array<string, 5> five{"12345", "23456", "34567", "45678", "56789"};
int main() {
string a, b;
while (cin >> a >> b){
bool flag = false;
if (b.size() <= 4){
while (b[0] < '9'){
for (char & c : b) c += 1;
if (a.find(b) != string::npos){
flag = true;
break;
}
}
}else{
a.resize(unique(a.begin(), a.end()) - a.begin()); //去重
for (int i = (b[0] - '0'); i < 5; ++i)
if (a.find(five[i]) != string::npos){
flag = true;
break;
}
}
if (flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
发表于 2023-03-01 22:43:18
回复(0)
a=input()
b=input()
l=len(b)
if l==1:
x=int(b)
flag=0
for i in a:
if int(i)>x:
flag=1
break
if flag==0:
print('NO')
else:
print('YES')
elif l==2:
x=int(b[0])
flag=0
for i in a:
if int(i)>x and a.count(i)>=2:
flag=1
break
if flag==0:
print('NO')
else:
print('YES')
elif l==3:
x=int(b[0])
flag=0
for i in a:
if int(i)>x and a.count(i)>=3:
flag=1
break
if flag==0:
print('NO')
else:
print('YES')
elif l==4:
x=int(b[0])
flag=0
for i in a:
if int(i)>x and a.count(i)>=4:
flag=1
break
if flag==0:
print('NO')
else:
print('YES')
else:
if b=='56789':
print('NO')
elif b=='45678':
if '5' in a and '6' in a and '7' in a and '8' in a and '9' in a:
print('YES')
else:
print('NO')
elif b=='34567':
if '5' in a and '6' in a and '7' in a and '8' in a :
if '4' in a&nbs***bsp;'9' in a:
print('YES')
else:
print('NO')
else:
print('NO')
elif b=='23456':
if '5' in a and '6' in a and '7' in a:
if ('3' in a and '4' in a)&nbs***bsp;('4' in a and '8' in a)&nbs***bsp;('8' in a and '9' in a):
print('YES')
else:
print('NO')
else:
print('NO')
else:
if '5' in a and '6' in a:
if ('2' in a and '3' in a and '4' in a)&nbs***bsp;('3' in a and '4' in a and '7' in a)&nbs***bsp;('4' in a and '7' in a and '8' in a) &nbs***bsp;('7' in a and '8' in a and '9' in a) :
print('YES')
else:
print('NO')
else:
print('NO')
发表于 2022-05-09 18:12:19
回复(0)
#include<stdio.h>
int main(){
char hand[105],desk[10];
while(scanf("%s",&hand)!=EOF){
int nums1[10],len2=0;
scanf("%s",&desk);
for(int i=0;i<10;i++){
nums1[i]=0;
}
for(int i=0;hand[i]!='\0';i++){
int x=hand[i]-'0';
nums1[x]++;
}
for(int i=0;desk[i]!='\0';i++){
len2++;
}
if(len2==1){
int flag=0,y=desk[0]-'0';
for(int i=y+1;i<10;i++){
if(nums1[i]!=0){
flag=1;
printf("YES");
break;
}
}
if(flag==0) printf("NO");
}
if(len2==2){
int flag=0,y=desk[0]-'0';
for(int i=y+1;i<10;i++){
if(nums1[i]>1){
flag=1;
printf("YES");
break;
}
}
if(flag==0) printf("NO");
}
if(len2==3){
int flag=0,y=desk[0]-'0';
for(int i=y+1;i<10;i++){
if(nums1[i]>2){
flag=1;
printf("YES");
break;
}
}
if(flag==0) printf("NO");
}
if(len2==4){
int flag=0,y=desk[0]-'0';
for(int i=y+1;i<10;i++){
if(nums1[i]>3){
flag=1;
printf("YES");
break;
}
}
if(flag==0) printf("NO");
}
if(len2==5){
int flag=0,y=desk[0]-'0';
for(int i=y+1;i<6;i++){
int count=0;
for(int j=0;j<5;j++){
if(nums1[i+j]!=0) count++;
}
if(count==5){
flag=1;
printf("YES");
break;
}
}
if(flag==0) printf("NO");
}
}
return 0;
}
发表于 2022-03-28 15:56:38
回复(0)
#include<stdio.h>
#include<string.h>
int main(void) {
int tmp[10] = {0};
char a[100];
char b[5];
scanf("%s", &a);
scanf("%s", &b);
int i;
for (i=0; i<strlen(a); i++) { //记录每个数字出现了多少次。
tmp[a[i]-'0']++;
}
if (strlen(b)<5) { // 非同花顺
for (i=b[0]-'0'+1; i<10 && tmp[i]<strlen(b);i++);
i==10?printf("NO"):printf("YES");
} else { // 同花顺
for (i=b[0]-'0'+1; i<6 && (tmp[i]*tmp[i+1]*tmp[i+2]*tmp[i+3]*tmp[i+4]==0); i++);
i==6?printf("NO"):printf("YES");
}
return 0;
}
发表于 2021-12-17 21:09:38
回复(0)
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