acm_最长子序列

题目：

[align=left]Problem Description[/align]
Given a sequence a[1],a[2],a[3]......a
, 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>

[align=left]Input[/align]
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>

[align=left]Output[/align]
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>

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 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。。