Problem 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
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思路：刚做过的洛谷原题。具体思路：点击这里
Source Program
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 21
#define MOD 123
#define E 1e-6
using namespace std;
int n,sum;
int triangle[N][N];
int a[N],vis[N];
bool flag;
void dfs(int step,int cnt)
{
if(flag)
return;
if(cnt>sum)//当前和大于sum，剪枝
return;
if(step==n+1&&cnt==sum)//达到最后一层并找到答案
{
flag=true;
for(int i=1;i<=n;i++)
cout<<a[i]<<" ";
}
for(int i=1;i<=n;i++)
if(!vis[i])
{
a[step]=i;
vis[i]=1;
cnt+=triangle[n][step]*a[step];//加上i*系数
dfs(step+1,cnt);
cnt-=triangle[n][step]*a[step];//减去i*系数
vis[i]=0;
}
}
int main()
{
cin>>n>>sum;
/*构造存储答案系数的杨辉三角*/
triangle[1][1]=1;
for(int i=2;i<=n;i++)
for(int j=1;j<=i;j++)
triangle[i][j]=triangle[i-1][j-1]+triangle[i-1][j];
dfs(1,0);//从1开始搜索
return 0;
}
