acm_最长子序列

题目：

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.<br>

 

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).<br>

 

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.<br>

 

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

题意：给你一个序列，然后求连续的一段数相加最大，求出这个最长子序列。。

想法：第一个用动态规划的题，用搜索应该也能做，但时间长，假定Max(k)表示以ak 做为“终点”的最长上升子序列的长度，那么：Max(1) = 1，Max (k) = Max { Max (i)：1<i < k 且 ai < ak 且 k≠1 } + 1。。然后累加值即可。。

代码：

#include <iostream>
#include<stdio.h>
using namespace std;
int main()
{
    //freopen("r.txt","r",stdin);
    int i,ca=1,t,s,e,n,x,now,before,max;
    cin>>t;
    while(t--)
    {
       cin>>n;
       for(i=1;i<=n;i++)
       {
         cin>>now;
         if(i==1)
         {
            max=before=now;
            x=s=e=1;
         }
         else {
             if(now>now+before)
             {
                before=now;
                x=i;
             }      
             else before+=now;
              }
         if(before>max)
           max=before,s=x,e=i;
       }
       cout<<"Case "<<ca<<":"<<endl;
       cout<<max<<" "<<s<<" "<<e<<endl;
       ca++;
       if(t!=0)
           cout<<endl;
    }
    return 0;
}

搜索比较容易想，代码比较长。。动态规划却相反。。