Robbers

Description

N robbers have robbed the bank. As the result of their crime they chanced to get M golden coins. Before the robbery the band has made an agreement that after the robbery i-th gangster would get Xi=Y of all money gained. However, it turned out that M may be not divisible by Y.

The problem which now should be solved by robbers is what to do with the coins. They would like to share them fairly. Let us suppose that i-th robber would get Ki coins. In this case unfairness of this fact is |Xi/Y - Ki/M|. The total unfairness is the sum of all particular unfairnesses. Your task as the leader of the gang is to spread money among robbers in such a way that the total unfairness is minimized.

 

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

 

Input

 

The first line of the input file contains numbers N, M and Y (1 <= N <= 1000, 1 <= M, Y <= 10000). N integer numbers follow - Xi (1 <= Xi <= 10000, sum of all Xi is Y).

 

 

Output

 

Output N integer numbers - Ki (sum of all Ki must be M), so that the total unfairness is minimal.

 

 

Sample Input

 

1

3 10 4
1 1 2

 

 

Sample Output

 

2 3 5

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#define ll long long
#define mod 1000000007
#include <bits/stdc++.h>
using namespace std;
typedef struct Dt
{
	int sum;
    double x;
}Dt;
/*int cmp( const void *a ,const void *b)
{
return (*(Dt *)a).x < (*(Dt *)b).x ? 1 : -1;
}*/
bool cmp(Dt a,Dt b)
{
    return a.x>b.x;
}
Dt a[2000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,y;
        scanf("%d %d %d",&n,&m,&y);
        int m1=m,y1=y;
        int b[2000]={0};
        for(int i=1;i<=n;i++)
        {
            double x;
            a[i].sum=i;
            scanf("%lf",&x);
            b[i]=(x*1)*m/y;
            m1-=b[i];
            a[i].x=x/(y*1.0)*(m*1.0)-b[i]*1.0+1e-6;
            //printf("x=%.3lf b[%d]=%d a[%d].x=%.3lf\n",x,i,b[i],i,a[i].x);
        }
        //qsort(a,n,sizeof(a[1]),cmp);
        sort(a+1,a+n+1,cmp);
        /*for(int i=1;i<=n;i++)
        {
            printf("a[%d].sum=%d a[%d].x=%.3lf\n",i,a[i].sum,i,a[i].x);
        }*/
        for(int i=1;i<=n;i++)
        {
            if(m1>0)
            {
                b[a[i].sum]++;
                m1--;
            }
            else break;
        }
        printf("%d",b[1]);
        for(int i=2;i<=n;i++)
        {
            printf(" %d",b[i]);
        }
        printf("\n");
    }
}