Max Sum hdu 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 115321    Accepted Submission(s): 26749

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

 

 

 

 

这个题目本来今天是没想写的，结果今天上课时候，老师讲了最长递增子序列的问题，有两个for ,三个for 的暴力办法，最后讲了分治的方法，让我觉得十分佩服，由此我想起了这个最长递增子序列的问题，DP是最好的办法，我很佩服哎，套用DP的模板也就是了

 

 

#include<stdio.h>int main(){    int i,j,k,c,n,temp,max,t;    int a[120055];    scanf("%d",&c);    k=0;    for(k=1;k<=c;k++)    {        //k++;        scanf("%d",&n);        for(i=1; i<=n; i++)            scanf("%d",&a[i]);        int start=1,end=1;        temp=a[1];        max=a[1];        int t=0;        int p=1;        for(i=2; i<=n; i++)        {            if(temp<0)            {                temp=0;                p=i;            }            temp=temp+a[i];            if(temp>max)            {                max=temp;                end=i;                start=p;            }        }        if(k==c)        {            printf("Case %d:\n",k);        printf("%d %d %d\n",max,start,end);        }        else        {        printf("Case %d:\n",k);        printf("%d %d %d\n\n",max,start,end);        }    }    return 0;}