poj 1722 SUBTRACT

SUBTRACT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2677		Accepted: 1182		Special Judge

Description

We are given a sequence of N positive integers a = [a1, a2, ..., aN] on which we can perform contraction operations.
One contraction operation consists of replacing adjacent elements ai and ai+1 by their difference ai-a_i+1. For a sequence of N integers, we can perform exactly N-1 different contraction operations, each of which results in a new (N-1) element sequence.

Precisely, let con(a,i) denote the (N-1) element sequence obtained from [a1, a2, ..., aN] by replacing the elements ai and a_i+1 by a single integer ai-a_i+1 :

con(a,i) = [a1, ..., ai-1, ai-a_i+1, a_i+2, ..., aN]
Applying N-1 contractions to any given sequence of N integers obviously yields a single integer.
For example, applying contractions 2, 3, 2 and 1 in that order to the sequence [12,10,4,3,5] yields 4, since :

con([12,10,4,3,5],2) = [12,6,3,5]

con([12,6,3,5]   ,3) = [12,6,-2]

con([12,6,-2]    ,2) = [12,8]

con([12,8]       ,1) = [4]

Given a sequence a1, a2, ..., aN and a target number T, the problem is to find a sequence of N-1 contractions that applied to the original sequence yields T.

Input

The first line of the input contains two integers separated by blank character : the integer N, 1 <= N <= 100, the number of integers in the original sequence, and the target integer T, -10000 <= T <= 10000.
The following N lines contain the starting sequence : for each i, 1 <= i <= N, the (i+1)^st line of the input file contains integer ai, 1 <= ai <= 100.

Output

Output should contain N-1 lines, describing a sequence of contractions that transforms the original sequence into a single element sequence containing only number T. The ith line of the output file should contain a single integer denoting the i^th contraction to be applied.
You can assume that at least one such sequence of contractions will exist for a given input.

Sample Input

5 4
12
10
4
3
5

Sample Output

2
3
2
1

Source

CEOI 1998

 

难是真的不难，一开始看到了这种操作，感觉后效性处理只能够用区间dp

然后我考虑了一下合并顺序的影响，突然发现只是+和-的区别，那就是一个非常明显的线性dp了

但是我不知道为什么感觉做不了，看了一下，结果不会设计状态
现在想想其实就很明显，很经典。。
感觉我的dp的方法不是那么明显

主要还是看经验，这个状态设计和背包是一类的

现在想想我会的状态设计也就那几种
一个是背包一类的，一个是区间dp一类，一个是LCS还有f[i]表示第i个选上
一个是树形dp的，一个是用一个维度来表示当前这个数字选不选，一个是环形的2倍范围
一个是传纸条的那种，也就是现在到了哪里，但是比较长

按我现在学到了状压dp来算的话应该就这些，

 

代码

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <map>
#include <string.h>
#define ll long long
using namespace std;
inline int read() {
    char c=getchar();int a=0,b=1;
    for(;c<'0'||c>'9';c=getchar())if(c=='-')b=-1;
    for(;c>='0'&&c<='9';c=getchar())a=a*10+c-48;return a*b;
}
int n,a[101],f[101][30001],T;
int main()
{
//    freopen(".in","r",stdin);
//    freopen(".out","w",stdout);
    n=read();T=read();
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
    }
    f[1][a[1]+10000]=1;
    f[2][a[1]-a[2]+10000]=-1;
    for(int i=3;i<=n;i++)
    {
        for(int j=0;j<=20000;j++)
        {
            if(f[i-1][j]!=0)
            {
                f[i][j+a[i]]=1;
                f[i][j-a[i]]=-1;
            }
        }
    }
    int ans[1001]={};int now=T+10000,cnt=0;    
    for(int i=n;i>=2;i--)
    {
        ans[i]=f[i][now];
        if(ans[i]==-1)now+=a[i];
        else if(ans[i]==1)now-=a[i];
    }
    for(int i=2;i<=n;i++)
    {
        if(ans[i]==1)
        {
            cout<<i-cnt-1<<endl;
            cnt++;
        }
    }
    for(int i=2;i<=n;i++)
    {
        if(ans[i]==-1)cout<<1<<endl;
    }
    return 0;
}