UVa 10339 - Watching Watches (数学)

题意

两个钟，一个慢a秒，一个慢b秒，问下一次重合的时候是几点。

思路

首先我们要算出下次重合的时候是第几天。

表的一圈是43200秒。一天是86400秒。

我们可以这么想象：假设a > b，b在a的前面43200秒处，那么要几天a才能追上b。

day = 43200 / (a-b)

追及问题，大家肯定都知道= =

算出了几天才能再次重合，我们就能得出在这几天A表实际走了多少秒。

passSeconds = day * (86400-a)

知道了走了多少秒就能知道现在表针的位置了。

先模一下圈数，得到秒数。然后依次得到分针和时针。

注意题目要输出最近的分钟。

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <cmath>//#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/hash_policy.hpp>using namespace std;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1#define FOR(i, a, b) for ((i)=(a); (i) < (b); (i)++)const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 2e4+10;const int MOD = 100007;const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };const int seed = 131;int cases = 0;typedef pair<int, int> pii;int main(){    //ROP;    //freopen("out.txt", "w", stdout);    int n, m;    while (~scanf("%d%d", &n, &m))    {        double day = 43200.0/abs(n-m);        LL slowTime = (LL)(day*(86400-n));        slowTime %= 43200;        int minute = slowTime / 60;        slowTime %= 60;        if (slowTime >= 30) minute++;        int hour = minute / 60;        minute %= 60;        if (hour == 0) hour = 12;        printf("%d %d %02d:%02d\n", n, m, hour, minute );    }    return 0;}