题目地址:点击打开链接
题意:先按时针和分针的角度排序,如何角度相等则按时间排序
思路:刚开始是想着时针和分针对0点的角度加起来,考虑的情况多且比较复杂,后来角度直接减取绝对值就可
m点n分k秒时
时针从0点转过的角度a = [m + n/60 + k/3600]*30
= 30m + n/2 + k/120(度)
分针转过的角度b = [n/60 + k/3600]*360
= 6n + k/10(度)
AC代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> typedef long long ll; using namespace std; struct node { int h,s; double r; }a[6]; bool cmp(struct node a,struct node b) { if(a.r != b.r) return a.r < b.r; else return a.h < b.h; } int main() { int t; int i; scanf("%d",&t); while(t--) { for(i=0; i<5; i++) { scanf("%d:%d",&a[i].h,&a[i].s); if(a[i].h >= 12) { a[i].r = abs(30.0 * (a[i].h-12) + a[i].s / 2.0 - a[i].s * 6.0); } else { a[i].r = abs(30.0 * a[i].h + a[i].s / 2.0 - a[i].s * 6.0); } if(a[i].r > 180.0) a[i].r = 360.0 - a[i].r; } sort(a,a+5,cmp); printf("%02d:%02d\n",a[2].h,a[2].s); } return 0; }
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> typedef long long ll; using namespace std; struct node { int h,s; double r; }a[6]; bool cmp(struct node a,struct node b) { if(a.r != b.r) return a.r < b.r; else return a.h < b.h; } int main() { int t; int i; scanf("%d",&t); while(t--) { for(i=0; i<5; i++) { scanf("%d:%d",&a[i].h,&a[i].s); if(a[i].h >= 12) { a[i].r = 30.0 * (a[i].h-12) + (60 - a[i].s) * 6.0 + 2.0 * a[i].s; } else { a[i].r = 30.0 * a[i].h + (60 - a[i].s) * 6.0 + 2.0 * a[i].s; } if(a[i].r > 180.0) a[i].r = 360.0 - a[i].r; } sort(a,a+5,cmp); printf("%02d:%02d\n",a[2].h,a[2].s); } return 0; }