poj1722 SUBTRACT【线性DP】

SUBTRACT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2037		Accepted: 901		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^thcontraction 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

题意：

给N个数，每次选定一个i，用a[i] - a[i+1]的结果取代a[i]和a[i+1]。操作N-1次之后，就只剩下1个数。问如何选择，可以使这个数恰好是T

思路：

我们可以发现，这个序列每次用减来取代的话，其实就是在n个数之前加上正负，然后让他们都加起来的和结果是T。

而且a[1]一定是正，a[2]一定是负。

于是我们用dp[i][j]来表示第i个数得到分数j时的符号，1表示是加，-1表示是减，0表示没有扩展到。【我怎样才能聪明地想到要这样来表示呢】

我们枚举i，j，判断上一个状态i-1有没有扩展到当前分数j。如果有，那么dp[i][j + a[i]] = 1, dp[i][j - a[i]] = -1

然后我们从dp[n][t]开始倒推出每一个元素之前的符号。ans[i]表示的就是a[i]前的符号。

那么方案该怎么输出？

先去处理所有的加号，相当于把所有的加号都挪去了原来a3的位置，这样最后处理减号的时候，减a2就可以负负得正了。

然后处理减号，把所有的减号都挪到了原来a2的位置，给a1来减。

 //#include <bits/stdc++.h>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<stdio.h>
 #include<cstring>
 #include<vector>
 #include<map>
 #include<set>

 #define inf 0x3f3f3f3f
 using namespace std;
 typedef long long LL;

 int n, t;
 int a[];
 int dp[][], ans[];
 const int base = ;

 int main()
 {
     while(scanf("%d%d", &n, &t) != EOF){
         for(int i = ; i <= n; i++){
             scanf("%d", &a[i]);
         }
         memset(dp, , sizeof(dp));
         dp[][a[] + base] = ;
         dp[][a[] - a[] + base] = -;
         for(int i = ; i <= n; i++){
             for(int j = - + base; j <=  + base; j++){
                 if(dp[i - ][j] != ){
                     dp[i][j + a[i]] = ;
                     dp[i][j - a[i]] = -;
                 }
             }
         }

         int want = t + base;
         for(int i = n; i >= ; i--){
             ans[i] = dp[i][want];
             if(ans[i] == ){
                 want -= a[i];
             }
             else if(ans[i] == -){
                 want += a[i];
             }
         }

         int cnt = ;
         for(int i = ; i <= n; i++){
             if(ans[i] == ){
                 printf("%d\n", i - cnt - );
                 cnt++;
             }
         }
         for(int i = ; i <= n; i++){
             if(ans[i] == -){
                 printf("1\n");
             }
         }
     }
     return ;
 }