poj 2940 Wine Trading in Gergovia
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1.题意:有n个居民买酒与卖酒,没移动一个房子距离相当用一个工作量,求最少工作量。
55 -4 1 -3 16-1000 -1000 -1000 1000 1000 10000
多个样例,每个样例有两行,5代表有5个居民,第二行代表每个居民的需求,整数代表买进,负数代表卖出;
2.思路:一家一家的访问,一家一家的累加即可;
3.优化后代码:
#include<stdio.h>int main(){ int n; int a; long long sum; long long ans; while(scanf("%d",&n)!=EOF) { if(n==0) break; sum=ans=0; for(int i=0;i<n;i++) { scanf("%d",&a); sum+=a; if(sum<0) ans-=sum; if(sum>=0) ans+=sum; } printf("%lld\n",ans); } return 0;}
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