Max Sum

拼搏现实的明天。 2022-08-18 02:30 136阅读 0赞

## Max Sum ##

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##### Time Limit: 2000ms Memory limit: 32768K 有疑问？点这里^\_^ #####

## 题目描述 ##

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.

## 输入 ##

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).

## 输出 ##

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.

## 示例输入 ##

2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5

## 示例输出 ##

Case 1:
    14 1 4
    
    Case 2:
    7 1 6

## 提示 ##

[hdoj1003][] 注意：本题后台测试数据量比较大，请使用C语言scanf输入。避免超时！！！！

## 来源 ##

Ignatius.L

#include<stdio.h>  
    int main()  
    {  
        int t;  
        scanf("%d",&t);  
        int cases=1;  
        while(t--)  
            {  
            int n,i;  
            scanf("%d",&n);  
            int sum=0,max=-9999;  
            int start=1,end=1,tmp=1;  
            int num;  
            for(i=1;i<=n;i++)  
            {  
                scanf("%d",&num);  
                sum+=num;  
                if(max<sum)  
                {  
                    max=sum;  
                    start=tmp;  
                    end=i;  
                }  
                if(sum<0)  
                {  
                    sum=0;  
                    tmp=i+1;  
                }  
            }  
            printf("Case %d:\n",cases);  
            printf("%d %d %d\n",max,start,end);  
            if(t!=0)  
                printf("\n");  
            cases++;  
        }  
        return 0;  
    }

[hdoj1003]: http://acm.hdu.edu.cn/showproblem.php?pid=1081