Backward Digit Sums（全排列next_permutation）

        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.       

题意：1.输入一个N，结果被限制为：只能是1到N的整数，而且只能有N个整数(即限定范围，又限定数量) 

    2.要求求出这N个数的排列，使他们按照类似杨辉三角型的规则进行相加后，结果为sum

题解：next_permutation()全排列函数的运用直至1~n中有一种排列使得最后结果为sum就结束 

#include<iostream>
#include<algorithm>
using namespace std;
int main(){
int n;
int sum;
int a[15];
int arr[15][15];
while(cin>>n>>sum){
for(int i=0;i<n;i++)
a[i]=i+1;
do{
for(int j=0;j<n;j++)
arr[0][j]=a[j];
for(int i=1;i<n;i++)
for(int j=0;j<n-i;j++)
arr[i][j]=arr[i-1][j]+arr[i-1][j+1];
if(arr[n-1][0]==sum)
break;
}while(next_permutation(a,a+n));
for(int i=0;i<n;i++)
cout<<a[i]<<" ";
cout<<endl;
}
return 0;
}