題目描述
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.
有這麼一個遊戲:
寫出一個1~N的排列a[i],然後每次將相鄰兩個數相加,構成新的序列,再對新序列進行這樣的操作,顯然每次構成的序列都比上一次的序列長度少1,直到只剩下一個數字位置。下面是一個例子:
3 1 2 4
4 3 6
7 9 16 最後得到16這樣一個數字。
現在想要倒著玩這樣一個遊戲,如果知道N,知道最後得到的數字的大小sum,請你求出最初序列a[i],為1~N的一個排列。若答案有多種可能,則輸出字典序最小的那一個。
[color=red]管理員注:本題描述有誤,這裡字典序指的是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]
輸入輸出格式
輸入格式:兩個正整數n,sum。
輸出格式:輸出包括1行,為字典序最小的那個答案。
當無解的時候,請什麼也不輸出。(好奇葩啊)
輸入輸出樣例
4 16
3 1 2 4
說明
對於40%的資料,n≤7;
對於80%的資料,n≤10;
對於100%的資料,n≤12,sum≤12345。
楊輝三角+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] = //先打一張楊輝三角表 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個數,值為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 }