HDOJ 1003 Max Sum（动态规划）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 176237    Accepted Submission(s): 41047

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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#include<stdio.h>#include<string.h>int main(){int n,i,a[100001],t=1,m;int thissum,maxsum,star,end,mark;//目前的和，历史最大和，最大和开始点，最大和终止点，标记（为定开始点用）scanf("%d",&n);while(n--){memset(a,0,sizeof(a));//初始化scanf("%d",&m);for(i=0;i<m;i++){scanf("%d",&a[i]);}thissum=0; maxsum=a[0]; star=1; end=1; mark=1;//从第一个数开始for(i=0;i<m;i++){thissum+=a[i];//目前和要加上当前数   if(thissum>maxsum){maxsum=thissum;star=mark;//其实mark不变，开始位置就不变 end=i+1;//结束位置是当前 }//发现历史最大和比当前和小了，最大连续子序列不唯一，则输出序号i和j最小的那个 ，所以不 =if(thissum<0){ mark=i+2;//标记后一个数，为star做准备   thissum=0; //当前最大值清理  }//目前最大和小于0，就不是最大了，要是等于0或大于0还好说  }printf("Case %d:\n%d %d %d\n",t++,maxsum,star,end);if(n) printf("\n");//格式 }return 0;}