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:

$d(A)=\max_{1\leq s_1\leq t_1<s_2\leq t_2\leq n}\left\{\sum_{i=s_1}^{t_1}a_i+\sum_{j=s_2}^{t_2}a_j\right\}$

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

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 = ;

int num[maxn];

int le[maxn], ri[maxn];

int main() {

    int T;

    scanf("%d", &T);

    while (T --) {

        memset(le, , sizeof(le));

        memset(ri, , sizeof(ri));

        int n;

        scanf("%d", &n);

        for (int i = ; i <= n; i ++) scanf("%d", &num[i]);

        le[] = num[];

        ri[n] = num[n];

        int cur_sum = num[] <  ?  : num[];

        for (int i = ; i <= n; i ++) {

            cur_sum += num[i];

            le[i] = max(le[i-], cur_sum);

            if (cur_sum < ) cur_sum = ;

        }

        cur_sum = num[n] <  ?  : num[n];

        for (int i = n - ; i >= ; i --) {

            cur_sum += num[i];

            ri[i] = max(ri[i+], cur_sum);

            if (cur_sum < ) cur_sum = ;

        }

        int ans = -0x3fffffff;

        for (int i = ; i <= n; i ++) {

            ans = max(ans, le[i-] + ri[i]);

        }

        printf("%d\n", ans);

    }

    return ;

}

巴特西

POJ2479（最长连续子序列和）

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