poj 3187 Backward Digit Sums 【全排列变形题】

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5356		Accepted: 3093

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

思路：

    输入n和sum,然后将1到n这n个数进行全排列，然后找出所有的排序中进行题上的计算之后等于sum的最小的字典序的那个序列！

代码：

方法1：（自己的方法）

#include 
#include 
#include 
using namespace std;
int n,sum;
int a[15];
int b[15];
int c[15];
int vis[15];
int flag;
void dfs(int pos,int n)
{
	if(flag==1)//剪枝！没有会超时 
		return;
	if(pos==n&&flag==0)
	{
		for(int i=0;i=1;j--)
		{
			for(int i=0;i


方法2：（利用全排列函数进行全排列）

#include 
#include 
#include 
using namespace std;
int N,sum;
int b[15];
int c[15];
int vis[15];
int flag;
void dfs(int pos,int n)
{
	do
	{
		for(int i=0;i=1;j--)
		{
			for(int i=0;i