HDU1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 190377    Accepted Submission(s): 44335

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

 

 

1.找到maxsum a[i] = max(maxsum a[i-1]+a[i],a[i])  这个规律。

2.注意在s<0情况下的处理。

3.DP的思路。

#include<iostream>
#include<cstdio>
int max(int x,int y )
{
    if (x>y)
        return x;
    else
        return y;
}
int main()
{
    int T,n,a,s,sum,x,y,z;
    while (~scanf("%d",&T))
    {
        for (int o=1;o<=T;o++)
        {
            scanf("%d",&n);
            z=1;s=0;sum=-1001;x=1,y=1;
            for (int p=1;p<=n;p++)
            {
                scanf("%d",&a);
                s+=a;
                if(s>sum)
                {
                    sum=s;
                    x=z;
                    y=p;
                }
                if(s<0)
                {
                    s=0;
                    z=p+1;
                }
            }
            printf("Case %d:\n",o);
            printf("%d %d %d\n",sum,x,y);
            if(o!=T)
                puts("");
        }
    }
    return 0;
}