ACDREAM 03C Robbers（贪心专场）

ACDREAM 03C Robbers

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 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.
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

3 10 4
1 1 2

Sample Output

2 3 5

题目大意：每组数据包含两个部分，第一行N， M， Y，第二行是N个整数X1 ～ XN。题目给出了一个公式：|Xi/Y−Ki/M|， 要求求出使这个式子的值的和最小的N个k的值（k的和要为M）。

解题思路：将|Xi/Y−Ki/M|转换成s[i]=(M∗Xi)/Y, 最后将s[i]相加得出sum，比较sum与M的大小，作出相应操作（具体见代码）。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#define N 1005using namespace std;typedef long long ll;int K[N];double S[N];struct rec{    int id;    double num;}r[N];int cmp1(rec a, rec b) {    return a.num < b.num;}int cmp2(rec a, rec b) {    return a.num > b.num;};int main() {    int n;    while (scanf("%d", &n) != EOF) {        int M, Y;        scanf("%d %d", &M, &Y);        int X, sum = 0;        for (int i = 0; i < n; i++) {            scanf("%d", &X);            S[i] = (M * X) / Y;            K[i] = (int)S[i];            if (1 - (S[i] - K[i]) < S[i] - K[i]) {                K[i]++;                     }            r[i].num = abs((S[i] + 1.0) / M - X * 1.0 / Y) - abs(S[i] * 1.0 / M - X * 1.0 / Y);            r[i].id = i;             sum += K[i];        }        if (sum < M) {            int temp = M - sum;            sort(r, r + n, cmp1);            for (int i = 0; i < temp; i++) {                K[r[i].id]++;            }        } else if (sum > M) {            int temp = sum - M;            sort(r, r + n, cmp2);            for (int i = 0; i < temp; i++) {                K[r[i].id]--;            }        }        printf("%d", K[0]);        for (int i = 1; i < n; i++) {            printf(" %d", K[i]);        }        printf("\n");    }    return 0;}