杭电ACM 1003 Max Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/u010935715/article/details/20453973

本文介绍了一种求解给定整数序列中最大子序列和的方法，包括输入格式、输出要求以及示例代码。

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简单题，题目意思就是求最大数，方法很简单

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 127953 Accepted Submission(s): 29640

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<stdio.h>
int main()
{
    int start,end,number,input,cas,max,ne,i,j,tempstart;
    scanf("%d",&cas);
    for(j=1;j<=cas;j++)
    {
        scanf("%d",&number);
        start=1;
        end=1;
        max=-99999999;
        tempstart=1;
        ne=0;
        
        for(i=1;i<=number;i++)
        {
            scanf("%d",&input);
            ne+=input;
            if(ne>max)
            {
                max=ne;
                start=tempstart;
                end=i;
            }
            if(ne<0)
            {
                tempstart=i+1;
                ne=0;
            }
        
        }
        
        printf("Case %d:\n%d %d %d\n",j,max,start,end);
        if(j!=cas)printf("\n");
    }
    
return 0;
}