【例题 8-5 UVA - 11054】Wine trading in Gergovia
2018-01-03 10:48
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【链接】 我是链接,点我呀:)
【题意】
在这里输入题意
【题解】
考虑第一个店。
如果它要酒的话,只能从第二个店那里运过来。
这样,问题就转化成后n-1个店的问题了。
然后会发现,第二家店它的情况也是同样的。
即a[2] = a[1]+a[2];
然后做相同的事情。a[2]的需求量只能从第3家店运过来(送过去)。
每一段路的花费就是∑ai的绝对值。
【代码】
/* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("") or putchar() or printf and such things? 5.init the used array or any value? 6.use error MAX_VALUE? 7.use scanf instead of cin/cout? 8.whatch out the detail input require */ /* 一定在这里写完思路再敲代码!!! */ #include <bits/stdc++.h> #define ll long long using namespace std; int n; int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_with_stdio(0),cin.tie(0); while (cin >> n && n){ ll now = 0,ans = 0; for (int i = 1;i <= n;i++){ int x; cin >> x; ans += abs(now); now+=x; } cout << ans << endl; } return 0; }
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