hdu 1003 Max Sum

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 <iostream>#include <cstring>#include <cstdio>#include<cmath>#include <algorithm>using namespace std;int nn=1; int star,end,f[100005],Max; void solve() {     int sum_star=1,sum=0; int flag=0;     star=end=1;Max=f[1];    for(int i=1;i<=f[0];i++)    {        if(f[i]>0) flag=1;      sum+=f[i];      if(sum>Max)      {          Max=sum; end=i;star=sum_star;      }     if(sum<0)     {         sum=0;         sum_star=i+1;     }    }   if(flag==0)   { Max=f[1]; end=star=1;     for(int i=1;i<=f[0];i++)       if(Max<f[i])     {        Max=f[i];        end=star=i;     }   } }int main(){    int T;  scanf("%d",&T);  while(T--)  {      scanf("%d",&f[0]);     for(int i=1;i<=f[0];i++)        scanf("%d",&f[i]);    solve();     printf("Case %d:\n",nn++);     printf("%d %d %d\n",Max,star,end);     if(T)        printf("\n");  }}