ZOJ 2343 Robbers

Robbers

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Time Limit: 5 Seconds      Memory Limit: 32768 KB      Special Judge

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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 X_i=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 K_i coins. In this case unfairness of this fact is |X_i/Y - K_i/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 - X_i (1 <= X_i <= 10000, sum of all X_i is Y).

Output

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

Sample Input

1

3 10 4
1 1 2

Sample Output

2 3 5

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Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #2

按比例放大xi，剩下的coin尽量分到可以使绝对值变小的强盗的手里。。。。。。。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

using namespace std;

int main()

{

    int T;

    scanf("%d",&T);

while(T--)

{

    int n,M,Y,a[1100],k[1100],res=0;

    double temp[1100];

    scanf("%d%d%d",&n,&M,&Y);

    for(int i=0;i<n;i++)

    {

        scanf("%d",&a);

        k=M/Y*a;

        temp=fabs(a*1.0/Y-k*1.0/M);

    }

    res=M-M/Y*Y;

    while(res--)

    {

        double minn=0x3f3f3f3f;

        int pos=-1;

        for(int i=0;i<n;i++)

        {

            if(minn>fabs(a*1.0/Y-(k*1.0+1)/M)-temp)

            {

                minn=fabs(a*1.0/Y-(k*1.0+1)/M)-temp;

                pos=i;

            }

        }

        k[pos]=k[pos]+1;

        temp[pos]=fabs(a[pos]*1.0/Y-k[pos]*1.0/M);

    }

    for(int i=0;i<n;i++)

        printf("%d ",k);

    putchar(10);

}

    return 0;

}