SDUT 2370 || POJ 2940 Wine Trading in Gergovia(简单贪心)
2012-03-25 02:44
399 查看
Wine Trading in Gergovia
As you may know from the comic “Asterix and the Chieftain’s Shield”, Gergovia consists of one street, and every inhabitant of the city is a wine salesman. You wonder how this economy works? Simple enough: everyone buys wine from other inhabitants of the city. Every day each inhabitant decides how much wine he wants to buy or sell. Interestingly, demand and supply is always the same, so that each inhabitant gets what he wants. There is one problem, however: Transporting wine from one house to another results in work. Since all wines are equally good, the inhabitants of Gergovia don’t care which persons they are doing trade with, they are only interested in selling or buying a specific amount of wine. They are clever enough to figure out a way of trading so that the overall amount of work needed for transports is minimized. In this problem you are asked to reconstruct the trading during one day in Gergovia. For simplicity we will assume that the houses are built along a straight line with equal distance between adjacent houses. Transporting one bottle of wine from one house to an adjacent house results in one unit of work. Input The input consists of several test cases. Each test case starts with the number of inhabitants n (2 ≤ n ≤ 100000). The following line contains n integers ai (−1000 ≤ ai ≤ 1000). If ai ≥ 0, it means that the inhabitant living in the ith house wants to buy ai bottles of wine, otherwise if ai < 0, he wants to sell −ai bottles of wine. You may assume that the numbers ai sum up to 0. The last test case is followed by a line containing 0. Output For each test case print the minimum amount of work units needed so that every inhabitant has his demand fulfilled. You may assume that this number fits into a signed 64-bit integer (in C/C++ you can use the data type “long long” or “__int64”, in JAVA the data type “long”). Sample Input 5 5 -4 1 -3 1 6 -1000 -1000 -1000 1000 1000 1000 0 Sample Output 9 9000 Source Ulm Local 2006 |
代码如下:
#include <iostream> #include <cstdio> using namespace std; int main() { int n, i; __int64 sum, ans, a; while (scanf("%d", &n) != EOF && n) { ans = 0; sum = 0; for (i = 0; i < n; ++i) { scanf("%I64d", &a); sum += a; if (sum > 0) { ans += sum; } else { ans -= sum; } } printf("%I64d\n", ans); } return 0; }
相关文章推荐
- POJ 2940 Wine Trading in Gergovia(简单贪心)
- POJ 2940 Wine Trading in Gergovia(简单贪心)
- poj 2940 Wine Trading in Gergovia 贪心+扫描
- UVa 11054/HDU 1489/POJ 2940 Wine trading in Gergovia(贪心&双向队列)
- UVa11054 poj2940 sdut2370 Wine trading in Gergovia(贪心)
- POJ 2940 Wine Trading in Gergovia【贪心】
- POJ-2940-Wine trading in Gergovia
- poj2940-Wine Trading in Gergovia
- POJ 2940 Wine Trading in Gergovia 可能会
- poj 1940 Wine Trading in Gergovia_贪心
- poj 1940 Wine Trading in Gergovia_贪心
- poj 2940 wine trading in gergovia
- Cpp环境 【poj 2940 】【Uva11054】【Vijos2909】Wine Trading in Gergovia 格尔高维亚的肮脏红酒交易
- poj 2940 Wine Trading in Gergovia
- 【每日一题(11)】Wine Trading in Gergovia POJ - 2940
- Wine trading in Gergovia uva+贪心
- Wine trading in Gergovia UVA - 11054 贪心思维
- UVA 11054 Wine trading in Gergovia 葡萄酒交易 贪心+模拟
- UVA-11054-Wine trading in Gergovia(模拟+贪心)
- Uva 11054 Wine trading in Gergovia(贪心模拟)