hdu 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 116194    Accepted Submission(s): 26942

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

Author

Ignatius.L

这个题目我做了很久，由于不会动归，所以请教别人才最后弄了出来；

分析：1  3  -5  -7  8

当遇到sum的和为<0，则sum=0;k=i+1,k的值赋值给初下标s1;s2则随循环移动记录，max值一定保持是最大的

#include<stdio.h>int main(){    int L,l,n,i,s1,s2,num[100005],max,sum;    scanf("%d",&L);    for(l=1;l<=L;l++)    {        scanf("%d",&n);        for(i=1;i<=n;i++)            scanf("%d",&num[i]);        max=-99999;        int k=1;        sum=0;        for(i=1;i<=n;i++)        {            sum+=num[i];            if(sum>max)            {                max=sum;                s1=k;                s2=i;            }            if(sum<0)            {                sum=0;                k=i+1;            }        }        printf("Case %d:\n",l);        if(l==L)  printf("%d %d %d\n",max,s1,s2);        else    printf("%d %d %d\n\n",max,s1,s2);    }    return 0;}