hdu 4442贪心

挺简单。假设我们现在面对两个要排的队a1, b1和a2, b2,已经花的时间为t,则先排1队需要的时间是a1+a2+t+tb1+tb2+tb1b2+a1b2,而先排2队需要时间是a1+a2+t+tb2+tb1+tb1b2+a2b1,所以实际上只要比较a1b2和a2b1的大小就能确定先排哪个队。按此排个序就行了。

/*
 * hdu4442/win.cpp
 * Created on: 2012-10-30
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
typedef long long LL;
const int MOD = 365 * 24 * 60 * 60;
inline bool cmp(const pair<int, int> &p1, const pair<int, int> &p2) {
    LL t1 = (LL)p1.first * p2.second;
    LL t2 = (LL)p2.first * p1.second;
    return t1 < t2;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int n, a, b;
    vector<pair<int, int> > que;
    while(scanf("%d", &n) == 1 && n > 0) {
        que.clear();
        for(int i = 0; i < n; i++) {
            scanf("%d%d", &a, &b);
            que.push_back(make_pair(a, b));
        }
        sort(que.begin(), que.end(), cmp);
        LL ans = 0;
        for(int i = 0; i < n; i++) {
            ans += (ans * (que[i].second % MOD) + (que[i].first % MOD)) % MOD;
            ans %= MOD;
        }
        printf("%d\n", (int)(ans % MOD));
    }
    return 0;
}
posted @ 2012-10-30 10:55  moonbay  阅读(...)  评论(... 编辑 收藏