Max Sum(hd P1003)

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

超时代码

#include<stdio.h>
int main()
{
    int a[100000],T,N,T1,j,i;
    scanf("%d",&T);
    T1=T;
    while(T--)
    {
        int Msum=0,sum=0,s=0,w=0;
        printf("case %d:\n",T1-T);
        scanf("%d",&N);
        for(i=0;i<N;i++)
            scanf("%d",&a[i]);
        Msum=a[0];
        for(j=0;j<N;j++)
        {
            for(i=j;i<N;i++)
            {
                if(a[i]<=0)
                {
                    sum+=a[i];
                    continue;
                }
                sum+=a[i];
                if(Msum<sum)
                {
                    s=j;
                    w=i;
                    Msum=sum;
                }
            }
            sum=0;
        }
        printf("%d %d %d\n\n",Msum,s+1,w+1);
    }
    return 0;
}

 

AC代码

/*状态转移方程 d[i] = max(d[i-1]+a[i], a[i])
　　d[i]表示以i位置结束的最大子序列之和。*/
#include<stdio.h>
int main()
{
    int a[100000];
    int T,T1;
    scanf("%d",&T);
    T1=T;
    while(T--)
    {
        int  sum=0,msum=0,i,x=0,y=0,start=0,end=0,N;
        scanf("%d",&N);
        for(i=0;i<N;i++)
            scanf("%d",&a[i]);
        sum=a[0];
        msum=sum;
        for(i=1;i<N;i++)
        {
            if(sum<0)/*dp[i-1]对a[i]不仅没有贡献，反而有损害，就应该舍弃*/
            {
                x=y=i;
                sum=a[i];
            }
            else
            {
                sum+=a[i];
                y=i;
            }
            if(sum>msum)
            {
                msum=sum;
                start=x;
                end=y;
            }        
        }
        printf("Case %d:\n",T1-T);
    
        if(T==0)
        printf("%d %d %d\n",msum,start+1,end+1);
        else
        printf("%d %d %d\n\n",msum,start+1,end+1);
        
    }
}