题目大意：每行输入三个数是一棵树的坐标，直到三个零结束。三个数当作坐标，求其他树到他的最小距离.
解题思路：两个循环向下取整（直接int强转）为0~9则，数组下标+1。数组往大开，距离用double存
ac代码：
#include <iostream> #include <cstring> #include <math.h> using namespace std; int dis(double a[], double b[]) { double sum; sum = pow( (a[0]-b[0]), 2 ) + pow( (a[1]-b[1]), 2 ) + pow( (a[2]-b[2]), 2 ); sum = (int)sqrt(sum); return sum; } void count(double a[][3], int n) { int b[10]; double temp, min; memset(b, 0, sizeof(b)); for (int i=0; i<n; i++){ min = dis(a[i], a[i+1]); for (int j=0; j<n; j++) if (j == i) continue; else{ temp = dis(a[i], a[j]); if (min > temp) min = temp; } if (min < 10) b[(int)min]++; } for (int i=0; i<10; i++) printf("M", b[i]); cout << endl; } int main() { double a[5005][3]; int i=0; while (scanf("%lf%lf%lf", &a[i][0], &a[i][1], &a[i][2])!=EOF){ if ( !a[i][0] && !a[i][1]&& !a[i][2] ) break; i++; } count(a, i); return 0; }