POJ:Maximum sum
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
t1 t2
d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }
i=s1 j=s2
Your task is to calculate d(A). - 输入
- The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case. - 输出
- Print exactly one line for each test case. The line should contain the integer d(A).
- 样例输入
-
1 10 1 -1 2 2 3 -3 4 -4 5 -5
- 样例输出
-
13
- 提示
- In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended. - 来源
- POJ Contest,Author:Mathematica@ZSU
#include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<vector> #include<cmath> #include<algorithm> const int INF = 0x3f3f3f3f; using namespace std; int main() { int T, n, data[50005]; int dp1[50005], dp2[50005]; int kk1[50005], kk2[50005]; scanf("%d", &T); while (T--){ scanf("%d", &n); for (int i = 1; i <= n; i++){ scanf("%d", &data[i]); } memset(dp1, 0, sizeof dp1); memset(dp1, 0, sizeof dp2); kk1[0] = -INF, kk2[n + 1] = -INF; for (int i = 1; i <= n; i++){ dp1[i] = max(dp1[i - 1] + data[i], data[i]); dp2[n - i + 1] = max(dp2[n - i + 2] + data[n - i + 1], data[n - i + 1]); kk1[i] = max(kk1[i - 1], dp1[i]); kk2[n - i + 1] = max(kk2[n - i + 2], dp2[n - i + 1]); } int ans = -INF; for (int i = 1; i <= n - 1; i++){ ans = max(ans, kk1[i] + kk2[i + 1]); } printf("%d\n", ans); } return 0; }
posted on 2018-12-03 22:58 zhouyuepku 阅读(212) 评论(0) 编辑 收藏 举报