简单的动态规划 Max Sum

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 

 

Sample Output

Case 1:14 1 4Case 2:7 1 6 

这个题很简单，但是！！自己开始做的时候不知道咋想的对起点的处理处理错了。。。其实很好想，如果前面那个dp[i-1]大于0，那么i的起点和i-1起点相同，否则i的起点就是自己所在位置，代码如下：

#include <stdio.h>#include <stdlib.h>int main(){    int t,n,i,ans,num[100000],dp[100000],st,end,cas,blankline;    blankline=cas=0;    int s[100000];    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(i=0;i<n;i++){            scanf("%d",&num[i]);        }        ans=dp[0]=num[0];        st=end=0;        s[0]=0;        for(i=1;i<n;i++){            if(dp[i-1]<0){                dp[i]=num[i];                s[i]=i;            }else{                dp[i]=dp[i-1]+num[i];                s[i]=s[i-1];            }        }        for(i=1;i<n;i++){            if(dp[i]>ans){                ans=dp[i];                st=s[i];                end=i;            }        }        if(blankline)            printf("\n");        printf("Case %d:\n",++cas);        printf("%d %d %d\n",ans,st+1,end+1);        blankline++;    }    return 0;}