题目描述
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.
有这么一个游戏:
写出一个1~N的排列a[i],然后每次将相邻两个数相加,构成新的序列,再对新序列进行这样的操作,显然每次构成的序列都比上一次的序列长度少1,直到只剩下一个数字位置。下面是一个例子:
3 1 2 4
4 3 6
7 9 16 最后得到16这样一个数字。
现在想要倒着玩这样一个游戏,如果知道N,知道最后得到的数字的大小sum,请你求出最初序列a[i],为1~N的一个排列。若答案有多种可能,则输出字典序最小的那一个。
[color=red]管理员注:本题描述有误,这里字典序指的是1,2,3,4,5,6,7,8,9,10,11,12
而不是1,10,11,12,2,3,4,5,6,7,8,9[/color]
输入输出格式
输入格式:两个正整数n,sum。
输出格式:输出包括1行,为字典序最小的那个答案。
当无解的时候,请什么也不输出。(好奇葩啊)
输入输出样例
4 16
3 1 2 4
说明
对于40%的数据,n≤7;
对于80%的数据,n≤10;
对于100%的数据,n≤12,sum≤12345。
杨辉三角+暴力!
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 int n,num; 7 const int PT[][13] = //先打一张杨辉三角表 8 { 9 { 1 } , // N = 1 10 { 1 , 1 } , // N = 2 11 { 1 , 2 , 1 } , // N = 3 12 { 1 , 3 , 3 , 1 } , // N = 4 13 { 1 , 4 , 6 , 4 , 1 } , // N = 5 14 { 1 , 5 , 10, 10, 5 , 1 } , // N = 6 15 { 1 , 6 , 15, 20, 15, 6 , 1 } , // N = 7 16 { 1 , 7 , 21, 35, 35, 21, 7 , 1 } , // N = 8 17 { 1 , 8 , 28, 56, 70, 56, 28, 8 , 1 } , // N = 9 18 { 1 , 9 , 36, 84,126,126, 84, 36, 9 , 1 } , // N = 10 19 { 1 , 10, 45,120,210,252,210,120, 45, 10 , 1 } , // N = 11 20 { 1 , 11, 55,165,330,462,462,330,165, 55 ,11 , 1 } }; // N = 12 21 int vis[15]; 22 int ans[15]; 23 int flag; 24 int dfs(int p,int k,int now)//第p个数,值为k,和是now 25 { 26 27 if(now>num)return 0; 28 if(p==n) 29 { 30 if(now==num) 31 { 32 ans[p]=k; 33 return 1; 34 } 35 else 36 return 0; 37 } 38 vis[k]=1; 39 for(int i=1;i<=n;i++) 40 { 41 if(vis[i]==0&&dfs(p+1,i,now+PT[n-1][p]*i)) 42 { 43 ans[p]=k; 44 return 1; 45 } 46 } 47 vis[k]=0; 48 return 0; 49 } 50 int main() 51 { 52 53 scanf("%d%d",&n,&num); 54 if(dfs(0,-1,0)) 55 for(int i=1;i<=n;i++) 56 printf("%d ",ans[i]); 57 return 0; 58 }