HDU 1003 Max Sum——经典dp之最大子序和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147102    Accepted Submission(s): 34379

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

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

 

Sample Output

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

 

#include<cstdio>#include<cstring>#include<algorithm>const int maxn=100000+10;using namespace std;int a[maxn],n;int main(){    int t,cas=0,sum,ans,begin,end,loc;     scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        sum=ans=a[1];        begin=end=loc=1;    //loc这个变量不能少，ＷＡ了好几次才找到                                //还有begin和end赋初值也不能丢        for(int i=2;i<=n;i++)        {            if(sum+a[i]<a[i])            {                sum=a[i];                loc=i;            }            else            {                sum+=a[i];            }            if(ans<sum)            {                ans=sum;                begin=loc;                end=i;            }        }        printf("Case %d:\n%d %d %d\n",++cas,ans,begin,end);        if(t)            printf("\n");    }    return 0;}