HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 243667    Accepted Submission(s): 57534

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

 

Author

Ignatius.L

 

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经典的动态规划（最长子序列）

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <map>#include <vector>#include <set>#define maxn 100010#define inf 0x3f3f3f3f3f3f#define mod 1000007using namespace std;typedef long long ll;int dp[maxn];int num[maxn];int main(){    int t;    int n;    scanf("%d",&t);    int cas=1;    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&num[i]);        }        int st=0,end=1;        ll maxsum=-100100;        ll sum=0;        int temp=1;        for(int i=1;i<=n;i++)        {            sum+=num[i];            if(sum>maxsum)            {                maxsum=sum;                st=temp;                end=i;            }            if(sum<0)            {                sum=0;                temp=i+1;            }        }        printf("Case %d:\n",cas++);        printf("%lld %d %d\n",maxsum,st,end);        if(t!=0)printf("\n");    }}