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题意:给定一个开始时间和一个角度,问你下一个时刻时针和分针形成这个角度是几点。
析:反正数量很小,就可以考虑暴力了,从第一秒开始暴力,直到那个角度即可,不会超时的,数目很少,不过要注意精度。
代码如下:
#include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> using namespace std ; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f3f; const double eps = 1e-8; const int mod = 360*120; const int maxn = 2000 + 5; const int dr[] = {0, 0, -1, 1}; const int dc[] = {-1, 1, 0, 0}; int n, m; inline bool is_in(int r, int c){ return r >= 0 && r < n && c < m && c >= 0; } int main(){ int h, m, s, t, kase = 0; while(scanf("%d:%d:%d", &h, &m, &s) == 3){ scanf("%d", &t); t *= 120; int sum = h * 3600 + m * 60 + s; int hh = sum % mod; int mm = (sum * 12) % mod; int ans = 0; while(true){ hh = (hh + 1) % mod; mm = (mm + 12) % mod; ++ans; if(abs(abs(mm-hh) - t) <= 10) break; } int sss = (ans + s) % 60; int mmm = ((ans+s) / 60 + m) % 60; int hhh = (((ans+s) / 60 + m) / 60 + h) % 12; printf("Case #%d: %02d:%02d:%02d\n", ++kase, hhh, mmm, sss); } return 0; }
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5719410.html