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[],T,N,T1,j,i;
     scanf("%d",&T);
     T1=T;
     while(T--)
     {
         int Msum=,sum=,s=,w=;
         printf("case %d:\n",T1-T);
         scanf("%d",&N);
         for(i=;i<N;i++)
             scanf("%d",&a[i]);
         Msum=a[];
         for(j=;j<N;j++)
         {
             for(i=j;i<N;i++)
             {
                 if(a[i]<=)
                 {
                     sum+=a[i];
                     continue;
                 }
                 sum+=a[i];
                 if(Msum<sum)
                 {
                     s=j;
                     w=i;
                     Msum=sum;
                 }
             }
             sum=;
         }
         printf("%d %d %d\n\n",Msum,s+,w+);
     }
     return ;
 }

 

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[];
     int T,T1;
     scanf("%d",&T);
     T1=T;
     while(T--)
     {
         int  sum=,msum=,i,x=,y=,start=,end=,N;
         scanf("%d",&N);
         for(i=;i<N;i++)
             scanf("%d",&a[i]);
         sum=a[];
         msum=sum;
         for(i=;i<N;i++)
         {
             if(sum<)/*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==)
         printf("%d %d %d\n",msum,start+,end+);
         else
         printf("%d %d %d\n\n",msum,start+,end+);

     }
 }