codeforces 339C 记忆化搜索

C. Xenia and Weights

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.

Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.

You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on ​​the scales or to say that it can't be done.

Input

The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).

Output

In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can putm weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.

If there are multiple solutions, you can print any of them.

Examples

input

00000001013

output

YES8 10 8

input

10000000002

output

NO

题意:  有一个天平, 有10种砝码, 分别往两边放置砝码, 每次放砝码必须满足两个条件: 一是放置的砝码不能跟上一次的一样, 二是放置这个砝码的时候, 这边的盘子的总重量必须大于另一边的总重量.
分析: 利用记忆化搜索来做就行了, 分别记录位置, 差值, 以及上一次的选择.

#include<bits/stdc++.h>#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;typedef pair<int,int> pii;const int N=1010,MOD=1e9+7;bool dp[N][12][12];int path[N];string s;int m;bool dfs(int pos,int differ,int pre){    if(pos>0 && (differ <=0 || differ > 10)) return 0;    if(pos == m){        puts("YES");        for(int i=0;i<m;i++) printf("%d ",path[i]);        return 1;    }    if(dp[pos][differ][pre]) return 0;    dp[pos][differ][pre]=1;    for(int i=1;i<=10;i++){        if(s[i-1]=='1' && i!=pre){            path[pos] = i;            if(dfs(pos+1,i-differ,i)) return 1;        }    }    return 0;}int main(){    cin>>s;    cin>>m;    if(!dfs(0,0,-1)) puts("NO");    return 0;}