POJ 2646 The Trip

POJ 2646 The Trip

题目链接：http://poj.org/problem?id=2646

题意：有n个学生的花费，计算使他们平摊花费所移动钱最少是多少。

注意：“the minimum amount of money that must change hands in order to equalize(within a cent) all the students' costs. ”

题目中最关键的一句，最少的钱——the minimum amount of money。精确到1美分——(within a cent)；

AC代码：

#include<stdio.h>int main(){int n,i;double a[1100],sum,ave;double ans1,ans2;while(scanf("%d",&n)!=EOF,n){sum=0;for(i=0;i<n;i++){scanf("%lf",&a[i]);sum+=a[i];}ave=((int)((sum/n+0.005)*100))/100.0;ans1=ans2=0;for(i=0;i<n;i++){if(a[i]>ave)ans1+=a[i]-ave;elseans2+=ave-a[i];}printf("$%.2lf\n",ans1>ans2? ans2:ans1);}return 0;}