poj3187Backward Digit Sums(dfs)

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Backward Digit Sums

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5673 Accepted: 3287

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:

    3   1   2   4      4   3   6        7   9         16

Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ's mental arithmetic capabilities.

Write a program to help FJ play the game and keep up with the cows.

Input

Line 1: Two space-separated integers: N and the final sum.

Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4

Hint

Explanation of the sample:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

//poj3187(深度优先搜索) //题目大意：给你两数n,m；问你从1~n的排列中是否存在一个序列；由该序列相邻两个数相加得到一个//新序列，不断如此变换直到最终为一个数；如果这个数等于m；并且在满足上述条件下字典序最小；//就输出该序列；//解体思路：就是用DFS按从小到大的顺序枚举1~n的排列；当满足上述条件就打印该序列停止搜索。 #include<cstdio>#include<cstring>int a[20],vis[20],n,sum,s,flag;bool jude(int sum)    //判断1~n的序列中是否能最终变换为m； {int b[20],j;int k=1;for(j=1;j<=n;j++) b[j]=a[j];  //将1~n的一个排列放到临时存放的数组b[]中 while(k<n){ j=1; for(int i=2;i<=n;i++) {b[j++]=b[i]+b[i-1];      //不断让序列中相邻两个数相加  } k++;    }    if(b[1]==sum) return true;    else          return false;}void dfs(int cur){if(flag==1) return ;if(cur==n+1&&jude(sum)&&flag==0)     //cur==n+1并且满足jude() {printf("%d",a[1]);for(int i=2;i<=n;i++) printf(" %d",a[i]);  //打印该序列 printf("\n");flag=1;                    //flag=1,停止搜索。 return ;}else{for(int i=1;i<=n;i++){a[cur]=i;               //从1到n枚举排列 if(!vis[i]){vis[i]=1;dfs(cur+1);vis[i]=0;}}}}int main(){int i,j,k;scanf("%d%d",&n,&sum);memset(vis,0,sizeof(vis));memset(a,0,sizeof(a));     //初始化a[],用来存放1~n的序列； flag=0;dfs(1);return 0;}

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