Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 251171    Accepted Submission(s): 59503

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

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

Author

Ignatius.L

Problem :     Judge Status : Accepted

RunId : 21239601    Language : G++    Author :

Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int INF = -(1<<30); int main(){ int T,n,cnt = 0; scanf("%d",&T); while(T--){ int ans ,start=1,tail,tmp = 0, x,ts; scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&x); if(i==1){ ans = tmp = x; start = tail = ts = 1; } else{ if(x > x + tmp){ tmp = x; ts = i; } else tmp = tmp + x; } if(tmp > ans ){ ans = tmp; start = ts,tail = i; } } printf("Case %d:\n%d %d %d\n",++cnt,ans,start,tail); if(T) printf("\n"); } }

#include
      
       #include
       
        #include
        
         using namespace std;const int INF = -(1<<30);int main(){
          int T,n,cnt = 0;    scanf("%d",&T);    while(T--){
              int ans ,start=1,tail,tmp = 0, x,ts;        scanf("%d",&n);        for(int i=1;i<=n;i++){
                  scanf("%d",&x);            if(i==1){
                      ans = tmp = x;                start = tail = ts = 1;            }            else{
                      if(x > x + tmp){
                          tmp = x;                    ts = i;                }                else tmp = tmp + x;            }            if(tmp > ans ){
                      ans = tmp;                start = ts,tail = i;            }        }        printf("Case %d:\n%d %d %d\n",++cnt,ans,start,tail);        if(T) printf("\n");    }}