题目：​​http://poj.org/problem?id=1722​​

SUBTRACT

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 1696

 

Accepted: 746

 

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-ai+1, ai+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

分析：很经典的dp题，开始真心想不出(甚至还往枚举搜索上想过||-_-，原来这是dp，现在发现动态规划在时间和空间上的好处真是巨大而明显)。


仿造的山寨代码：


/*
不断的作差，改变差顺序可以使得整个序列得到t。由于顺序的改变可能使得'-'变成'+'，
比如12-(10-4-(3-5))=12-(10-4-3)-5=12-10+4+3-5.
于是问题就变成了n个数字的加减组合得到t。即a[1]-a[2]±a[3]±a[4]±...±a[n]=t.
进一步转化：±a[3]±a[4]±...±a[n]=t-a[1]+a[2]. 背包问题！--》
2a[3]+2[4]+...+2a[n]=t-a[1]+a[2]+(a[3]+[4]+...+a[n]);  2a[i]可选可不选，不选对应减
*/
#include <cstdio>
#include <cstring>
#include <cstring>
#define V 30000
#define N 200

int number[N];
int dp[V];
int record[N][V];
bool tag[N];
int sta[N];

int main() {
    //freopen("cin.txt","r",stdin);
    int n,t;
    while (~scanf("%d%d", &n, &t)) {
        int a, b;
        if (n == 1) {
            scanf("%d", &a);
            continue;
        }
        if (n == 2) {
            scanf("%d%d", &a, &b);
            printf("%d\n", 1);
            continue;
        }
        scanf("%d%d", &a, &b);
        int sum = 0;
        for (int i = 0; i < n-2; i++) {
            scanf("%d", &number[i]);
            sum += number[i];
            number[i] *= 2;
        }
        int v = sum-a+b+t;
        memset(dp,0,sizeof(dp));
        for (int i = 0; i < n-2;i++) {  //物件数
            for (int j = v; j - number[i] >= 0; j--) {
                if (dp[j] > dp[j-number[i]]+number[i]) record[i][j] = 0;  //不取 对应着‘-’
                else {
                    dp[j] = dp[j-number[i]]+number[i];
                    record[i][j] = 1;  //取 ‘+’
                }
            }
        }
        //printf("dp = %d(v= %d)\n", dp[v], v);
        int nowv = v;
        for (int i = n-2-1; i >= 0; i--) {
            if (record[i][nowv] == 1) {
                tag[i] = 1;
                nowv -= number[i];  //等式两边减去加上的数字
            }
            else tag[i] = 0;
        }

        int top = 0;
        for (int i = 0; i < n-2; i++) {
            if (tag[i] == 1) sta[top++] = i+2;  //dex
        }
        for (int i = 0; i < top; i++) sta[i] -= i;  //很重要，i其实也就是加号个数-1
        for (int i = 0; i < top; i++)  printf("%d\n", sta[i]);
        for (int i = top; i < n-1; i++) puts("1");
    }
    return 0;
}