最大连续子序列 HDU

Given a sequence a11,a22,a33......ann, 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 <iostream>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <algorithm>#include <vector>#include <string>#include <cstring>#include <sstream>#define INF 100000000using namespace std;int a[100005];int main(){    int T;    int kase=1;    scanf("%d",&T);    while(T--)    {        if(kase!=1) printf("\n");        printf("Case %d:\n",kase++);        int tmp=0;        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int maxx=-9999999;        int b=0,e=0,l=0;        for(int i=0;i<n;i++)        {            tmp+=a[i];            if(tmp>maxx)            {                maxx=tmp;                e=i;                b=l;            }            if(tmp<0)            {                tmp=0;                l=i+1;            }        }        printf("%d %d %d\n",maxx,b+1,e+1);    }    return 0;}