ACdream 1224 Robbers （贪心）

Robbers

Special JudgeTime Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Problem 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 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.

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

3 10 41 1 2

Sample Output

2 3 5

Source

Andrew Stankevich Contest 2

Manager

mathlover

解析：先对每个人都按比例向下取整得到k[i]，之后再将剩下的金币每次都分给当前“不公平值”最大的那个人，这样就会让他的“不公平值”减小，直到所有的金币分配完为止。

AC代码：

#include <bits/stdc++.h>using namespace std;int k[1002], x[1002];bool flag[1002];int main(){    #ifdef sxk        freopen("in.txt", "r", stdin);    //    freopen("out.txt", "w", stdout);    #endif // sxk    int n, m, y, sum;    while(scanf("%d%d%d", &n, &m, &y)!=EOF){        sum = 0;        for(int i=0; i<n; i++){            scanf("%d", &x[i]);            k[i] = m * x[i] / y;            if(m * x[i] % y == 0) flag[i] = true;            sum += k[i];        }        int tx = m - sum;        while(tx--){            double t = 0;            int j = 0;            for(int i=0; i<n; i++){                if(flag[i]) continue;                double foo = (double)x[i]/y - (double)k[i]/m;                if(t < foo){                    t = foo;                    j = i;                }            }            k[j] ++;        }        for(int i=0; i<n-1; i++) printf("%d ", k[i]);        printf("%d\n", k[n-1]);    }    return 0;}