POJ2479（最长连续子序列和）

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 37035		Accepted: 11551

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:Mathematica@ZSU

 

题意：原串分成两个不相交连续子序列，求最大和

const int maxn = 50010;

int num[maxn];
int le[maxn], ri[maxn];

int main() {

    int T;
    scanf("%d", &T);
    while (T --) {
        memset(le, 0, sizeof(le));
        memset(ri, 0, sizeof(ri));
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; i ++) scanf("%d", &num[i]);
        le[1] = num[1];
        ri[n] = num[n];
        int cur_sum = num[1] < 0 ? 0 : num[1];
        for (int i = 2; i <= n; i ++) {
            cur_sum += num[i];
            le[i] = max(le[i-1], cur_sum);
            if (cur_sum < 0) cur_sum = 0;
        }
        cur_sum = num[n] < 0 ? 0 : num[n];
        for (int i = n - 1; i >= 1; i --) {
            cur_sum += num[i];
            ri[i] = max(ri[i+1], cur_sum);
            if (cur_sum < 0) cur_sum = 0;
        }
        int ans = -0x3fffffff;
        for (int i = 2; i <= n; i ++) {
            ans = max(ans, le[i-1] + ri[i]);
        }
        printf("%d
", ans);
    }

    return 0;
}