兰博教训提莫之后,然后和提莫讨论起约德尔人,谈起约德尔人,自然少不了一个人,那 就是黑默丁格------约德尔人历史上最伟大的科学家. 提莫说,黑默丁格最近在思考一个问题:黑默丁格有三个炮台,炮台能攻击到距离它小于等于R的敌人 (两点之间的距离为两点之间的直线距离,例如(3,0),(0,4)之间的距离是5),如果一个炮台能攻击 到敌人,那么就会对敌人造成1×的伤害.黑默丁格将三个炮台放在N*M方格中的点上,并且给出敌人 的坐标. 问:那么敌人受到伤害会是多大?
[编程题]炮台攻击
importjava.util.*;publicclassMain{publicstaticvoidmain(String[] args) {Scanner sc = newScanner(System.in);while(sc.hasNext()) {intr = sc.nextInt();intcount = 0;int[] x = newint[4];int[] y = newint[4];for(inti = 0; i < 4; i++) {x[i] = sc.nextInt();y[i] = sc.nextInt();}for(inti = 0; i < 3; i++) {if((x[i] - x[3])*(x[i] - x[3]) + (y[i] - y[3]) * (y[i] - y[3]) <= r * r) {count++;}}System.out.println(count + "x");}}}
发表于 2016-08-01 12:19:46
回复(0)
#include<iostream>
#include<cmath>
using namespace std;
int main(void){
int a[8];
int R,i;
double d1,d;
while(cin>>R){
int count =0;
for(i=0;i<8;i++){
int x;
cin>>x;
a[i]=x;
}
for(i=0;i<5;i=i+2){
d1=(pow((a[i]-a[6]),2) + pow((a[i+1]-a[7]),2));
d=sqrt(d1);
if(d <= R)
count+=1;
}
cout<<count<<"x"<<endl;
}
return 0;
}
通过所有测试用例
发表于 2016-03-22 16:10:45
回复(0)
#include<stdio.h>
#include<math.h>
#define DIS(x,y,xx,yy) (sqrt((xx-x)*(xx-x)+(yy-y)*(yy-y)))
int main(){
int x1,y1,x2,y2,x3,y3,x0,y0;
float R;
while(~scanf("%f%d%d%d%d%d%d%d%d",&R,&x1,&y1,&x2,&y2,&x3,&y3,&x0,&y0)){
int sum=0;
if(DIS(x1,y1,x0,y0)<=R) sum++;
if(DIS(x2,y2,x0,y0)<=R) sum++;
if(DIS(x3,y3,x0,y0)<=R) sum++;
printf("%dx\n",sum);
}
return 0;
}
发表于 2016-03-03 11:09:56
回复(0)
import java.util.Scanner;
import java.lang.Math;
public class Main {
static int hurt = 0;
public static void main(String[] args) {
System.out.println("请输入攻击距离R和三个炮台的坐标x1, y1, x2, y2, x3, y3,
敌人的坐标x0, y0");
Scanner in = new Scanner(System.in);
double R = in.nextInt();
double x1 = in.nextInt();
double y1 = in.nextInt();
double x2 = in.nextInt();
double y2 = in.nextInt();
double x3 = in.nextInt();
double y3 = in.nextInt();
double x0 = in.nextInt();
double y0 = in.nextInt();
injure(x1, y1, x0, y0, R);
injure(x2, y2, x0, y0, R);
injure(x3, y3, x0, y0, R);
System.out.println("总共受到的伤害是：" + hurt + "X");
}
public static int injure(double x, double y, double x0, double y0, double r) {
if (Math.sqrt(Math.pow(x - x0, 2) + Math.pow(y - y0, 2)) <= r) {
hurt++;
}
return hurt;
}
}
大家好，这是我做的答案，在eclipse里面测试牛客网给的测试用例能通过，但是在线给的测试用例通不过，有同感的吗？
评论里找到原因了：测试用例会有很多组，这个只能算一组，要用while（in.hasNext()）循环不断接收数据，这样的话每组算完要将hurt重置为0XX
编辑于 2016-03-24 16:29:14
回复(14)
import java.io.*; import java.util.*;//我是第一次在线练习笔试，看了大家的才知道不能随便取名。之前取名Test编译都通不过。public class Main{ public static void main(String args[]){ Scanner reader = new Scanner(System.in); int R,x1,y1,x2,y2,x3,y3,x0,y0; //System.out.println("请输入R、x1,y1,x2,y2,x3,y3,x0,y0");
//不能随便输出东西，只能输出结果 while(reader.hasNext()){ int sum = 0;//每轮数据初始化 R = reader.nextInt(); x1 = reader.nextInt(); y1 = reader.nextInt(); x2 = reader.nextInt(); y2 = reader.nextInt(); x3 = reader.nextInt(); y3 = reader.nextInt(); x0 = reader.nextInt(); y0 = reader.nextInt(); if (getSum(x1,y1,x0,y0,R) == 1){ sum++; } if(getSum(x2,y2,x0,y0,R) == 1){ sum++; } if(getSum(x3,y3,x0,y0,R) ==1){ sum++; } System.out.println(sum+"x"); } } private static int getSum(int x,int y, int m,int n,int r){ if( ((m-x)*(m-x) +(n-y)*(n-y)) <= (r*r ) ){ return 1;//在范围内返回1 } else{ return 0;//范围外返回0 } } }
发表于 2016-03-18 13:31:26
回复(0)
很简单，直接计算三个炮台与敌人之间的间距，如果距离在炮台的射程之内就增加伤害。唯一需要注意的是本题有多组用例，记得循环读入用例。
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
while((line = br.readLine()) != null){
String[] params = line.split(" ");
int R = Integer.parseInt(params[0]);
int x1 = Integer.parseInt(params[1]);
int y1 = Integer.parseInt(params[2]);
int x2 = Integer.parseInt(params[3]);
int y2 = Integer.parseInt(params[4]);
int x3 = Integer.parseInt(params[5]);
int y3 = Integer.parseInt(params[6]);
int x0 = Integer.parseInt(params[7]);
int y0 = Integer.parseInt(params[8]);
int count = 0;
if((x1 - x0)*(x1 - x0) + (y1 - y0)*(y1 - y0) <= R*R){
count++;
}
if((x2 - x0)*(x2 - x0) + (y2 - y0)*(y2 - y0) <= R*R){
count++;
}
if((x3 - x0)*(x3 - x0) + (y3 - y0)*(y3 - y0) <= R*R){
count++;
}
System.out.println(count + "x");
}
}
}
发表于 2022-01-13 13:00:47
回复(0)
def damage(R,x1,y1,x2,y2,x3,y3,x0,y0): count = 0 if (x0-x1)**2 + (y0-y1)**2<=R**2: count = count + 1 if (x0-x2)**2 + (y0-y2)**2<=R**2: count = count + 1 if (x0-x3)**2 + (y0-y3)**2<=R**2: count = count + 1 print(str(count)+"x") while True: try: R,x1,y1,x2,y2,x3,y3,x0,y0 = map(int,input().strip().split()) damage(R,x1,y1,x2,y2,x3,y3,x0,y0) except: break
发表于 2021-09-25 12:22:07
回复(0)
import math
def get_distance(att, enemy, r):
dis = math.sqrt((att[0] - enemy[0]) ** 2 + (att[1] - enemy[1]) ** 2)
return 1 if dis <= r else 0
while True:
try:
R, x1, y1, x2, y2, x3, y3, x0, y0 = map(int, input().split())
enemy = (x0, y0)
atts = [(x1, y1), (x2, y2), (x3, y3)]
res = 0
for att in atts:
res += get_distance(att, enemy, R)
print('{}x'.format(str(int(res))))
except:
break
发表于 2020-08-19 16:13:11
回复(0)
import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int R=sc.nextInt();
int[] x=new int[3];
int[] y=new int[3];
x[0]=sc.nextInt();
y[0]=sc.nextInt();
x[1]=sc.nextInt();
y[1]=sc.nextInt();
x[2]=sc.nextInt();
y[2]=sc.nextInt();
int x0=sc.nextInt();
int y0=sc.nextInt();
int count=0;
for(int i=0;i<3;i++){
if((x[i]-x0)*(x[i]-x0)+(y[i]-y0)*(y[i]-y0)<=R*R){
count++;
}
}
System.out.println(count+"x");
}
}
}
发表于 2018-10-15 14:26:57
回复(0)
package com.special.first;
import java.util.Scanner;
/**
* 网易02-炮台攻击
* 这是考察欧几里得距离公式吗？？
*
* Create by Special on 2018/3/11 16:12
*/
public class Pro063 {
static int[] row = new int[4], col = new int[4];
public static boolean isInRange(double R, int x1, int y1, int x2, int y2){
return Math.pow(x1 - x2, 2) + Math.pow(y1 - y2, 2) <= R;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
while(input.hasNext()){
double R = input.nextInt();
R *= R;
for(int i = 0; i < 4; i++){
row[i] = input.nextInt();
col[i] = input.nextInt();
}
int count = 0;
for(int i = 0; i < 3; i++){
if(isInRange(R, row[i], col[i], row[3], col[3])){
count++;
}
}
System.out.println(count + "x");
}
}
}
发表于 2018-03-11 16:24:48
回复(0)
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int R,x1,y1,x2,y2,x3,y3,x0,y0;
while(cin>>R>>x1>>y1>>x2>>y2>>x3>>y3>>x0>>y0)
{
int count =0;
double R1=sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0));
if(R1<R)
count++;
double R2=sqrt((x2-x0)*(x2-x0)+(y2-y0)*(y2-y0));
if(R2<R)
count++;
double R3=sqrt((x3-x0)*(x3-x0)+(y3-y0)*(y3-y0));
if(R3<R)
count++;
cout<<count<<"x"<<endl;
}
return 0;
}
发表于 2017-06-30 19:00:15
回复(1)
//到三个炮台分别到敌人的距离即可
import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String[] s= sc.nextLine().split(" ");
int R = Integer.parseInt(s[0]);
int[] x = new int[3];
int[] y = new int[3];
x[0] = Integer.parseInt(s[1]);
y[0] = Integer.parseInt(s[2]);
x[1] = Integer.parseInt(s[3]);
y[1] = Integer.parseInt(s[4]);
x[2] = Integer.parseInt(s[5]);
y[2] = Integer.parseInt(s[6]);
int x0 = Integer.parseInt(s[7]);
int y0 = Integer.parseInt(s[8]);
int count= 0;
for(int i=0;i<3;i++){//循环3次，分别算三个距离
if(Math.sqrt(Math.pow(x[i]-x0,2) + Math.pow(y[i]-y0,2)) <= R){
count++;
}
}
System.out.println(count+"x");
}
}
}
发表于 2017-03-09 16:57:47
回复(0)
//很朴素的思想，大家都能看懂的答案，通过了测试
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()) {
int count = 0;
int R = scanner.nextInt();
int x1 = scanner.nextInt();
int y1 = scanner.nextInt();
int x2 = scanner.nextInt();
int y2 = scanner.nextInt();
int x3 = scanner.nextInt();
int y3 = scanner.nextInt();
int x0 = scanner.nextInt();
int y0 = scanner.nextInt();
if (attack(x0, y0, R, x1, y1))
count++;
if (attack(x0, y0, R, x2, y2))
count++;
if (attack(x0, y0, R, x3, y3))
count++;
System.out.println(count + "x");
}
}
public static boolean attack(int x0, int y0, int R, int
x, int y) {
return Math.pow(x0 - x, 2) + Math.pow(y0 - y, 2) <=
Math.pow(R, 2);
}
}
发表于 2017-02-25 09:56:33
回复(0)
// 不需要sqrt()函数，直接平方求距离即可
#include <iostream>
using namespace std;
int main(){
int R, x1, y1, x2, y2, x3, y3, x0, y0;
while(cin >> R >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x0 >> y0){
int d12 = (x1 - x0)*(x1 - x0) + (y1 - y0)*(y1 - y0);
int d22 = (x2 - x0)*(x2 - x0) + (y2 - y0)*(y2 - y0);
int d32 = (x3 - x0)*(x3 - x0) + (y3 - y0)*(y3 - y0);
int count = 0;
if(d12 <= R*R)
count ++;
if(d22 <= R*R)
count ++;
if(d32 <= R*R)
count ++;
cout << count << "x" << endl;
}
return 0;
}
编辑于 2016-09-05 16:57:49
回复(0)
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int r;
while(cin>>r)
{
int pao[3][2];
for(int i=0;i<3;i++)
cin>>pao[i][0]>>pao[i][1];
int x0,y0;
cin>>x0>>y0;
int cc=0;
for(int i=0;i<3;i++)
{
float dist = sqrt(float(pao[i][0]-x0)*(pao[i][0]-x0)+float(pao[i][1]-y0)*(pao[i][1]-y0));
if(dist <= r) cc++;
}
cout<<cc<<"x"<<endl;
}
}
发表于 2016-08-18 18:00:32
回复(0)
// 直观题，没任何算法，只是初中数学
#include<iostream>
#include<cmath>
using namespace std;
int main(void)
{
int R, x1, y1, x2, y2, x3, y3, x0, y0;
int hurt = 0;
while ( cin >> R >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x0 >> y0 )
{
if ( R >= sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0)) )
++hurt;
if ( R >= sqrt((x2-x0)*(x2-x0)+(y2-y0)*(y2-y0)) )
++hurt;
if ( R >= sqrt((x3-x0)*(x3-x0)+(y3-y0)*(y3-y0)) )
++hurt;
cout << hurt << 'x' << endl;
hurt = 0;
}
return 0;
}
编辑于 2016-08-16 22:09:52
回复(0)
#include <iostream>
#include <math.h>
using namespace std;
int check(int a, int b, int x, int y, int distant) {
float dis;
dis = sqrt((x-a)*(x-a) + (y-b)*(y-b));
if(dis <= (float)distant) return 1;
else return 0;
}
int main(){
int Data[8];
int i,R,max;
while(cin>>R){
max=0;
for(i=0; i<8; i++)
cin>>Data[i];
for(i=0; i<=4; i+=2)
max+=check(Data[i], Data[i+1], Data[6], Data[7], R);
cout<<max<<"x"<<endl;
}
return 0;
}
编辑于 2016-08-02 23:11:23
回复(0)
#include<iostream>#include<cmath>#include<vector>using namespace std;intmain(){intR,x1,y1,x2,y2,x3,y3,x0,y0;while(cin>>R>>x1>>y1>>x2>>y2>>x3>>y3>>x0>>y0){doubledis;intcnt = 0;doubleR1 = sqrt(pow((x1-x0),2)+pow((y1-y0),2));doubleR2 = sqrt(pow((x2-x0),2)+pow((y2-y0),2));doubleR3 = sqrt(pow((x3-x0),2)+pow((y3-y0),2));vector<double> vr;vr.push_back(R1);vr.push_back(R2);vr.push_back(R3);for(inti = 0;i<vr.size();++i)if(vr[i]<=R)++cnt;cout<<cnt<<'x'<<endl;}return0;}
发表于 2016-08-01 16:34:30
回复(0)
问题信息
通过挑战的用户
查看代码-
2023-01-05 17:42:36
-
2023-01-04 00:08:10 -
2022-09-15 09:22:17
-
2022-09-10 17:36:55 -
2022-09-09 22:22:03
相关试题
-
扫描二维码，关注牛客网
- 意见反馈
-
下载牛客APP，随时随地刷题
扫一扫，把题目装进口袋
- 求职之前，先上牛客
-
扫描二维码，进入QQ群
扫描二维码，关注牛客公众号
- 公司地址：北京市朝阳区北苑路北美国际商务中心K1座一层-北京牛客科技有限公司
- 联系方式：010-60728802 投诉举报电话：010-57596212（朝阳人力社保局）
- 牛客科技© All rights reserved admin@nowcoder.com
- 京ICP备14055008号-4 增值电信业务经营许可证 营业执照 人力资源服务许可证
-
京公网安备
11010502036488号
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; inline int sqr(int x){return x * x;} int main(){ int R,x[6],y[6]; while(scanf("%d",&R) != EOF){ for(int i = 0;i < 4;++ i) scanf("%d%d",&x[i],&y[i]); int ret = 0; for(int i = 0;i < 3;++ i) if(sqr(x[i] - x[3]) + sqr(y[i] - y[3]) <= sqr(R)) ++ ret; printf("%dx\n",ret); } return 0; }