C 解法, 执行用时: 1ms, 内存消耗: 328KB, 提交时间: 2021-09-06

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#include "math.h"
int main() {
	int R, x[4], y[4];
	int i,count;
	while (~scanf("%d %d %d %d %d %d %d %d %d", &R, &x[1], &y[1], &x[2], &y[2],&x[3],&y[3], &x[0], &y[0])) {
		count = 0;
		for (i = 1; i < 4; i++) {
			if ((x[i] - x[0])*(x[i] - x[0])+(y[i]-y[0])*(y[i]-y[0]) <= R * R)
				count++;
		}
		printf("%dx\n", count);
	}
	return 0;
}

C++ 解法, 执行用时: 1ms, 内存消耗: 372KB, 提交时间: 2017-07-24

/*#include <stdio.h>

int main() 
{
	int r, x1, y1, x2, y2, x3, y3, x0, y0;
	scanf("%d%d%d%d%d%d%d%d%d",
		&r, &x1, &y1, &x2, &y2, &x3, &y3, &x0, &y0);
	int t = 0;
#define SQR(x) ((x)*(x))
#define TEST(x, y) do{\
	if ((SQR(x - x0) + SQR(y - y0)) <= SQR(r)) t++;\
}while(0)
	TEST(x1, y1);
	TEST(x2, y2);
	TEST(x3, y3);
    printf("%dx\n", t);
	return 0;
}*/
#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;
}

C 解法, 执行用时: 2ms, 内存消耗: 304KB, 提交时间: 2021-09-21

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double dst(int x0, int y0, int x1, int y1)
{
	double res;
	double x;
	x = (x0 - x1) * (x0 - x1) + (y0 - y1) * (y0 - y1);
	res = sqrt(x);
	return res;
}

int main()
{
	int x0, y0, x1, y1, x2, y2, x3, y3, R;
	double t;
	int res = 0;
	scanf("%d %d %d %d %d %d %d %d %d", &R, &x1, &y1, &x2, &y2, &x3, &y3,
		  &x0, &y0);
	do {
		t = dst(x0, y0, x1, y1);
		if(t - R <= 0.000001)
			res += 1;
		t = dst(x0, y0, x2, y2);
		if(t - R <= 0.000001)
			res += 1;
		t = dst(x0, y0, x3, y3);
		if(t - R <= 0.000001)
			res += 1;
		printf("%dx\n",res);
		res = 0;
	}while(scanf("%d %d %d %d %d %d %d %d %d", &R, &x1, &y1, &x2, &y2, &x3, &y3,
		&x0, &y0) != EOF);
}

C 解法, 执行用时: 2ms, 内存消耗: 320KB, 提交时间: 2021-09-21

#include<stdio.h>

int main()
{
    int R;
    int arr[4][2];
    while(~scanf("%d%d%d%d%d%d%d%d%d",&R,&arr[0][0],&arr[0][1],&arr[1][0],&arr[1][1],&arr[2][0],&arr[2][1],&arr[3][0],&arr[3][1]))
    {
        int sum=0;
        for(int i=0;i<3;i++)
        {
            int distance=(arr[i][0]-arr[3][0])*(arr[i][0]-arr[3][0])+(arr[i][1]-arr[3][1])*(arr[i][1]-arr[3][1]);
            if(distance<=R*R){
                sum++;
            }
        }
        printf("%dx\n",sum);
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 324KB, 提交时间: 2022-05-10

#include <stdio.h>

int main()
{
    int pao[8];
    int R;
    while(scanf("%d",&R)!=EOF){
    for(int i=0;i<8;i++)
        scanf("%d",&pao[i]);
    
    int cnt=0;
    for(int i=0;i<6;i+=2)
    {
        if((pao[i]-pao[6])*(pao[i]-pao[6])+(pao[i+1]-pao[7])*(pao[i+1]-pao[7])<=R*R)
            cnt++;
        //printf("%d",(pao[i]-pao[6])*(pao[i]-pao[6])+(pao[i+1]-pao[7])*(pao[i+1]-pao[7]));
    }
    printf("%dx\n",cnt);
    }
    return 0;
}