HDU 1003 Max Sum (动规)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 141547    Accepted Submission(s): 32929

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

 

Author

Ignatius.L

 

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此题与HDU 1231一样，只是简单一些，记录前标要easy。

#include <iostream>
using namespace std;
#define M 100100
int vis[M],dp[M];
int main(int i,int j,int k)
{
    int n,last,t,first,temp;
    cin>>t;
    for(k=1;k<=t;k++)
    {
        cin>>n;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++) {cin>>vis[i];dp[i]=vis[i];}
        int    sum=0,maxSum=vis[1];first=last=temp=1;
        for(i = 1; i <= n; i++)
        {
            sum += vis[i];          //i是从1開始的，所以先加再推断。

if(sum > maxSum){       
                maxSum = sum;
                first = temp;
                last = i;
            }
            if(sum < 0){            //这里用sum表示vis[first]->vis[i]的和，所以要清零。

sum = 0;
                temp = i+1;
            }
        }
        printf("Case %d:\n",k);
        printf("%d %d %d\n",maxSum,first,last);
        if(k < t)
            printf("\n");  好吧，这里就让我格式错误了几次。。。明明Case 1:和Case 2:数据中间有空行，可是完毕一次没空行。

。

。
 }
    return 0;
}