Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 324393 Accepted Submission(s): 77146

Problem Description

　　Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

　　The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

　　For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:

14 1 4

Case 2:

7 1 6

【题意】

　　给定一个序列a[1]，a[2]，a[3]，.....，a[n]，来计算子序列的最大和。例如，给定序列(6，-1，5，4，-7)，这个序列中的最大和是 6 +(-1)+ 5 + 4 = 14。

　　输入的第一行包含一个整数T(1<=T<=20)，它表示测试用例的数量。然后是T行，每一行以N开头(1<=N<=100000)，然后是N个整数(所有整数都在-1000到1000之间)

　　对于每个测试用例，输出两行。第一行是“Case #:”，#表示测试用例的数量。第二行包含三个整数，序列中的最大和，子序列的起始位置，子序列的结束位置。如果有多个结果，输出第一个。在两种情况之间输出空行。

【代码】

#include <stdio.h>

#include <stdlib.h>

int main(){

    int i,j,n,t;

    scanf("%d",&t);

    for(i=;i<=t;i++){

        int *a,first=,last=,sum=,tmp=,max=-;

        scanf("%d",&n);

        a=(int*)malloc(n*sizeof(int));

        for(j=;j<n;j++){

            scanf("%d",&a[j]);

            sum += a[j];

            if(sum>max){

                max=sum;

                first=tmp;

                last=j+;

            }

            if(sum<){

                sum=;

                tmp=j+;

            }

        }

        printf("Case %d:\n%d %d %d\n",i,max,first,last);

        if(i!=t)printf("\n");

        a=NULL;

    }

    return ;

}

巴特西

HDU1003 最大连续子序列

Max Sum

【题意】

【代码】

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