hdu1003——Max Sum（在线算法+求起点终点）-白红宇

hdu1003——Max Sum（在线算法+求起点终点）

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发布时间：2019-05-10

本文共 1500 字，大约阅读时间需要 5 分钟。

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

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

对在线算法有了新的认识，两个if的顺序居然都不能错。。。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #define INF 0x3f3f3f3f#define MAXN 110000#define Mod 10001using namespace std;int main(){ int t,cnt=1; scanf("%d",&t); while(t--) { int n,m,k=1; int ans=-INF,tmp=0,ansi,ansj; scanf("%d",&n); for(int i=1; i<=n; ++i) { scanf("%d",&m); tmp+=m; if(tmp>ans) { ansi=k; ansj=i; ans=tmp; } if(tmp<0) { tmp=0; k=i+1; } } if(cnt!=1) printf("\n"); printf("Case %d:\n",cnt++); printf("%d %d %d\n",ans,ansi,ansj); } return 0;}

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