2016.3.20 H/E - Backward Digit Sums（ next_permutation写全排列）

呀吧。这个题居然是这样子的，自己补得，笑哭，就是一个全排列，因为数据太小所以不会超时

全排列加上递归求树的和， next_permutation用来求全排列十分方便

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.

代码，真好玩，这是我的代码，等会看看大神的代码，是不是会简单

#include 
#include
using namespace std;
int n,s;
int cr[15];
int sum[15];
int arr(int x)
{
    if(x==1)
        return 1;
    else
        return x*arr(x-1);

}
int ju(int x)
{
    if(x==1)
    {
        if(sum[1]==s)
            return 1;
        else
            return 0;
    }
    for(int i=1,j=1;i>n>>s;
    for(int i=1;i<=n;i++)
       cr[i]=i;
    for(int i=1;i<=arr(n);i++)
    {
        for(int j=1;j<=n;j++)
            sum[j]=cr[j];
        if(ju(n))
            break;
        next_permutation(cr+1,cr+n+1);
    }
    for(int i=1;i

差不多

不过他们求和的时候用的不是一个递归，而是用的两个for循环

int ju()
{
    for(int i=n;i>1;i--)
        for(int j=1;j

不错，不错，然后去看了看lyj的代码，哇真简洁，可棒get到了（虽然思路是一样的可是人家总是写的那么简洁）

#include 
#include
using namespace std;
int n,s;
int cr[15];
int sum[15];
int main()
{
    cin>>n>>s;
    for(int i=1;i<=n;i++)
       cr[i]=i;
       int x=n;
    while(x!=1)
    {
        for(int i=1;i<=x;i++)sum[i]+=sum[i+1];
        x--;
        if(x==1)
        {
            if(sum[1]==s)
                break;
            else
            {
                x=n;
                next_permutation(cr+1,cr+n+1);
                for(int i=1;i<=n;i++)
                    sum[i]=cr[i];

            }
        }
    }
    for(int i=1;i