Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204790    Accepted Submission(s): 47879

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

#include<iostream>
#include<cstdio>
using namespace std; 
int main() 
{ 
    int j,i,k,n,m,t; 
    int a;  //不需要数组，只需要一个输入变量
    scanf(“%d”,&t);
    for (j=1;j<=t;j++)
    {
        scanf(“%d”,&n);
        int sum=0,maxsum=-10000002,first =0, last = 0, temp = 1;
        for (i=0;i<n;i++)
        {
            scanf(“%d”,&a);
            sum += a;
            if (sum > maxsum)
            {
                maxsum = sum;first = temp;last = i+1;
            }
            if (sum < 0)
            {
                sum = 0;temp = i+2;
            }
        }
        printf(“Case %d:\n%d %d %d\n”,j,maxsum,first,last);
        if (j!=t) 
        { 
            printf(“\n”); 
        } 
    } 
     
    return 0; 
}  

最大字串和：没有输出最大字串和的未知的代码：

#include<iostream>
#include<cstdio>
using namespace std; 
int main() 
{ 
    int j,i,k,n,m,t; 
    int a;  //不需要数组，只需要一个输入变量
    scanf(“%d”,&t);
    for (j=1;j<=t;j++)
    {
        scanf(“%d”,&n);
        int sum=0,maxsum=-10000002;
        for (i=0;i<n;i++)
        {
            scanf(“%d”,&a);
            sum += a;
            if (sum > maxsum)
            {
                maxsum = sum;;
            }
            if (sum < 0)
            {
                sum = 0;;
            }
        }
        printf(“Case %d:\n%d \n”,j,maxsum);
        if (j!=t) 
        { 
            printf(“\n”); 
        } 
    }  
    return 0; 
}