Backward Digit Sums --- DFS+全排列

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.

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

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.

4 16

3 1 2 4

Explanation of the sample:

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

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;
}