Backward Digit Sums --- DFS+全排列

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4946 Accepted: 2850

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.

Source

USACO 2006 February Gold & Silver


解题分析 + 代码：

//基本的思路枚举所有的n的数字的组合，//当遇到和等于m的时候就输出。//这里有一个处理就是每次都是从最小的开始枚举，//因此所选出来的也就是按照字典序最小的，#include<iostream>#include<algorithm>#include<cstdio>using namespace std;//把读入的数字进行处理inline int read(){int x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}int a[15], f[15];int n, m;int main(){n = read();m = read();//初始化数组，设定大小for(int i = 1; i <= n; i++)a[i] = i;//全排列do{//现将值赋给f[];for(int i = 1; i <= n; i++)f[i] = a[i];//接着求杨辉三角的方式，依次两两相加for(int i = 1; i < n; i++)for(int j = 1; j <= n - i; j++)f[j] += f[j + 1];//判断是否等于所给定的结果if(f[1] == m){for(int i = 1; i < n; i++)cout<<a[i]<<" ";cout<<a[n]<<endl;return 0;}}while(next_permutation(a + 1,a + n + 1));return 0;}