hdu 1275纯数学题

找出足够的相遇或追及的时间，排个序输出就可以了。

/*

 * hdu1275/win.cpp

 * Created on: 2012-10-23

 * 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 struct Node{

    double t;

    double dis;

    Node() {

        dis = t = 0;

    }

    Node(double tt, double dd) {

        t = tt;

        dis = dd;

    }

}Node;

inline bool operator<(const Node &n1, const Node &n2) {

    return n1.t < n2.t;

}

int main() {

#ifndef ONLINE_JUDGE

    freopen("data.in", "r", stdin);

#endif

    int T, n;

    double S, v1, v2, v3, v4;

    scanf("%d", &T);

    while(T--) {

        set<Node> myset;

        scanf("%lf%lf%lf%d", &S, &v1, &v2, &n);

        v3 = v1 + v2;

        v4 = fabs(v1 - v2);

        for(int k = 1; k < 3000; k += 2) {

            double t = k * S / v3;

            double dis = t * v1;

            dis = dis - (int)(dis / S) * S;

            if(2 * dis > S) {

                dis = S - dis;

            }

            myset.insert(Node(t, dis));

            t = k * S / v4;

            dis = t * v1;

            dis = dis - (int)(dis / S) * S;

            if(2 * dis > S) {

                dis = S - dis;

            }

            myset.insert(Node(t, dis));

        }

        set<Node>::iterator it = myset.begin();

        while(n-- > 1) {

            it++;

        }

        printf("Time=%.3f Dist=%.3f\n", (*it).t, (*it).dis);

    }

    return 0;

}