hdu 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 245162    Accepted Submission(s): 57896

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
题目解析：
这是一个简单的DP题，类似于最大上升子序列问题，我们只要可以求出最大值就可以输出开始和结尾了
代码：
#include <iostream>#include<cstdio>using namespace std;int main(){    int j,i,k,n,m,t;    int a[100002];    scanf("%d",&t);    for (j=1;j<=t;j++)    {        scanf("%d",&n);        for (i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int sum=0,maxsum=-1001,first =0, last = 0, temp = 1;        for (i=0;i<n;i++)        {            sum += a[i];            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;}