Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 34697 | Accepted: 10752 |
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).
Input
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.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1 10 1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
Source
POJ Contest,Author:[email protected]
求两段的最大和,前后进行两次dp;
#include<stdio.h> #include<iostream> #include<math.h> #include<stdlib.h> #include<ctype.h> #include<algorithm> #include<vector> #include<string> #include<queue> #include<stack> #include<set> #include<map> using namespace std; int t, n, p[100010], ans[100010]; int main() { scanf("%d",&t); while (t--) { memset(ans,0,sizeof(ans)); scanf("%d",&n); int b = 0, sum = -1000000000; for (int i = 0; i < n; i++) { scanf("%d", &p[i]); b += p[i]; sum = max(b, sum); ans[i] = sum; if (b < 0) b = 0; } b = 0, sum = -1000000000; int tmp = -1000000000; for (int i = n - 1; i >= 1; i--) { b += p[i]; sum = max(b, sum); tmp = max(tmp, ans[i - 1] + sum); if (b < 0) b = 0; } printf("%d\n",tmp); } return 0; }
时间: 2024-08-29 12:17:21