BZOJ1653: [Usaco2006 Feb]Backward Digit Sums

1653: [Usaco2006 Feb]Backward Digit Sums

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 207 Solved: 161
[Submit][Status][Discuss]

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

OUTPUT DETAILS:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4
is the lexicographically smallest.

HINT

Source

Silver

题解：

暴力枚举排列是必须的，那我们看一下能否在O(N)的时间内检验一个排列按题意操作之后和是否为final number

根据直觉每一个数被使用的次数应该有某种关系，从样例来看：

7 9

4 3 6

3 1 2 4

他们的次数三角形为：

1 1

1 2 1

1 3 3 1

这下应该就明显了吧，下面的数的使用次数等于它们斜上角的两个数的使用之和，这不就是喜闻乐见的杨辉三角吗，组合数搞定。

代码：

待UPD

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 15

 #define maxm 500+100

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 #define for0(i,n) for(int i=0;i<=(n);i++)

 #define for1(i,n) for(int i=1;i<=(n);i++)

 #define for2(i,x,y) for(int i=(x);i<=(y);i++)

 #define for3(i,x,y) for(int i=(x);i>=(y);i--)

 #define mod 1000000007

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }

 int a[maxn],n,m,c[maxn][maxn];

 int main()

 {

     freopen("input.txt","r",stdin);

     freopen("output.txt","w",stdout);

     n=read();m=read();

     for1(i,n)

      {

          c[i][]=c[i][i]=;

         for1(j,n-)c[i][j]=c[i-][j]+c[i-][j-];

      }

     for1(i,n)a[i]=i; 

     while()

     {

         int tmp=;

         for1(i,n)tmp+=a[i]*c[n][i-];

         if(tmp==m)break;

         next_permutation(a+,a+n+);

     } 

     printf("%d",a[]);

     for2(i,,n)printf(" %d",a[i]);

     return ;

 }