Design T-Shirt

Soon after he decided to design a T-shirt for our Algorithm Board on Free-City BBS, XKA found that he was trapped by all kinds of suggestions from everyone on the board. It is indeed a mission-impossible to have everybody perfectly satisfied. So he took a poll to collect people's opinions. Here are what he obtained: N people voted for M design elements (such as the ACM-ICPC logo, big names in computer science, well-known graphs, etc.). Everyone assigned each element a number of satisfaction. However, XKA can only put K (<=M) elements into his design. He needs you to pick for him the K elements such that the total number of satisfaction is maximized. 

Input

The input consists of multiple test cases. For each case, the first line contains three positive integers N, M and K where N is the number of people, M is the number of design elements, and K is the number of elements XKA will put into his design. Then N lines follow, each contains M numbers. The j-th number in the i-th line represents the i-th person's satisfaction on the j-th element. 

Output

For each test case, print in one line the indices of the K elements you would suggest XKA to take into consideration so that the total number of satisfaction is maximized. If there are more than one solutions, you must output the one with minimal indices. The indices start from 1 and must be printed in non-increasing order. There must be exactly one space between two adjacent indices, and no extra space at the end of the line. 

Sample Input

3 6 42 2.5 5 1 3 45 1 3.5 2 2 21 1 1 1 1 103 3 21 2 32 3 13 1 2

Sample Output

6 5 3 12 1

题意：有n个人和m件衣服,给出每个人对m件衣服的评价，输出评价最大的前k件衣服,要按下标从大到小输出；

这个题的意思理解起来确实有点困难，主要有两种要求，第一是对衣服本身的要求（评价最高，并且编号要小），第二是对输出的要求（编号从大到小输出）

#include<stdio.h>#include<algorithm>using namespace std;struct s{    int num;    double mark;} a[1000],t[1000];bool cmp(s x,s y){    if(x.mark!=y.mark)        return x.mark>y.mark;    return x.num<y.num;}bool cmp1(s x,s y){   return x.num>y.num;}int main(){    int n,m,k,i,j,o,s;    double d;    while(~scanf("%d %d %d",&n,&m,&k))    {        for(i=0; i<m; i++)            a[i].mark=0;        while(n--)        {            for(i=0; i<m; i++)            {                scanf("%lf",&d);                a[i].mark+=d;                a[i].num=i+1;            }        }        sort(a,a+m,cmp);//先按总评价从大到小,相同时按编号从小到大        for(i=0;i<k;i++)            t[i]=a[i];//将前k个存起来        sort(a,a+k,cmp1);//再按编号从大到小输出        for(i=0; i<k-1; i++)        {            printf("%d ",a[i].num);        }        printf("%d\n",a[i].num);    }    return 0;}