UVA 11054 Wine trading in Gergovia
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题目大意:给出一组序列,每个数字代表一家店,正数表示需要多少酒,负数表示要卖多少酒,将k的酒从到相邻店需要k个劳动力,最少多少劳动力可以供需平衡
解题思路:遍历序列,将数字清0,得到数字代表清0需要的劳动力和下家店需要增加或者减少的酒,依次相加即可
#include <iostream>using namespace std;int n;int a[200000];long long int result;int main() { while(cin >> n && n) { result = 0; int last = 0; for(int i = 0; i < n; i++) { cin >> a[i]; last = a[i] + last; result+= abs(last); } cout << result << endl; } return 0;}
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- UVa 11054 - Wine trading in Gergovia
- UVa 11054 - Wine trading in Gergovia
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- UVA 11054 - Wine trading in Gergovia
- UVA 11054 Wine trading in Gergovia
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