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

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int a[100000],T,N,T1,j,i;
 5     scanf("%d",&T);
 6     T1=T;
 7     while(T--)
 8     {
 9         int Msum=0,sum=0,s=0,w=0;
10         printf("case %d:\n",T1-T);
11         scanf("%d",&N);
12         for(i=0;i<N;i++)
13             scanf("%d",&a[i]);
14         Msum=a[0];
15         for(j=0;j<N;j++)
16         {
17             for(i=j;i<N;i++)
18             {
19                 if(a[i]<=0)
20                 {
21                     sum+=a[i];
22                     continue;
23                 }
24                 sum+=a[i];
25                 if(Msum<sum)
26                 {
27                     s=j;
28                     w=i;
29                     Msum=sum;
30                 }
31             }
32             sum=0;
33         }
34         printf("%d %d %d\n\n",Msum,s+1,w+1);
35     }
36     return 0;
37 }

 

AC代码

 1 /*状态转移方程 d[i] = max(d[i-1]+a[i], a[i])
 2 　　d[i]表示以i位置结束的最大子序列之和。*/
 3 #include<stdio.h>
 4 int main()
 5 {
 6     int a[100000];
 7     int T,T1;
 8     scanf("%d",&T);
 9     T1=T;
10     while(T--)
11     {
12         int  sum=0,msum=0,i,x=0,y=0,start=0,end=0,N;
13         scanf("%d",&N);
14         for(i=0;i<N;i++)
15             scanf("%d",&a[i]);
16         sum=a[0];
17         msum=sum;
18         for(i=1;i<N;i++)
19         {
20             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*/
21             {
22                 x=y=i;
23                 sum=a[i];
24             }
25             else
26             {
27                 sum+=a[i];
28                 y=i;
29             }
30             if(sum>msum)
31             {
32                 msum=sum;
33                 start=x;
34                 end=y;
35             }        
36         }
37         printf("Case %d:\n",T1-T);
38     
39         if(T==0)
40         printf("%d %d %d\n",msum,start+1,end+1);
41         else
42         printf("%d %d %d\n\n",msum,start+1,end+1);
43         
44     }
45 }