Wine trading in Gergovia UVA
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很简单的一道题目,关键在于意识到每个酒庄最终剩余的数量均为0,那么从第一个酒庄的酒也就需要不断的向外转移,那么需要工作量也就是第一个酒庄藏酒量的绝对值(可以将藏酒量看作为负值),那么从第一个酒庄转移到第二个酒庄就继续向后转移,这里能够保证最后一个酒庄一定满足为0的条件,所以特殊处理了一下,具体实现见如下代码:
#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;int n;typedef long long LL;int main(){while (cin >> n){if (n == 0) break;LL ans = 0;int last = 0;for (int i = 0; i < n-1; i++){int t;cin >> t;last += t;ans += abs(last);}cin >> last;cout << ans << endl;}}
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