UVA10026

题意：裁缝有N个鞋子要修理，一天只能修一双鞋子，每个鞋子有规定修的天数以及每延迟一天修要罚钱的数目，要求找出花费最少罚钱的顺序思路：贪心，fine = money / day;fine越大，就要求越早修理，使得总的罚钱最少#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;struct job{    int num;    double day, money, fine;}j[1005];int cmp(job a, job b) {    if (fabs(a.fine - b.fine) > 0.000001)        return a.fine > b.fine;    return a.num < b.num;}int main() {    int cas;    scanf("%d", &cas);    while (cas--) {        int n;         scanf("%d", &n);        for (int i = 0; i < n; i++) {            scanf("%lf %lf", &j[i].day, &j[i].money);                   j[i].fine = j[i].money / j[i].day;            j[i].num = i + 1;        }         sort(j, j + n, cmp);        for (int i = 0; i < n - 1; i++)            printf("%d ", j[i].num);        printf("%d", j[n - 1].num);        if (cas)            printf("\n\n");        else            printf("\n");    }    return 0;}