POJ 2167 Irrelevant Elements

Description

Young cryptoanalyst Georgie is investigating different schemes of generating random integer numbers ranging from 0 to m - 1. He thinks that standard random number generators are not good enough, so he has invented his own scheme that is intended to bring more randomness into the generated numbers. 
First, Georgie chooses n and generates n random integer numbers ranging from 0 to m - 1. Let the numbers generated be a1, a2, . . . , an. After that Georgie calculates the sums of all pairs of adjacent numbers, and replaces the initial array with the array of sums, thus getting n - 1 numbers: a1 + a2, a2 + a3, . . . , a_n-1 + an. Then he applies the same procedure to the new array, getting n - 2 numbers. The procedure is repeated until only one number is left. This number is then taken modulo m. That gives the result of the generating procedure. 
Georgie has proudly presented this scheme to his computer science teacher, but was pointed out that the scheme has many drawbacks. One important drawback is the fact that the result of the procedure sometimes does not even depend on some of the initially generated numbers. For example, if n = 3 and m = 2, then the result does not depend on a2. 
Now Georgie wants to investigate this phenomenon. He calls the i-th element of the initial array irrelevant if the result of the generating procedure does not depend on ai. He considers various n and m and wonders which elements are irrelevant for these parameters. Help him to find it out.

Input

Input contains n and m (1 <= n <= 100 000, 2 <= m <= 10⁹).

Output

On the first line of the output print the number of irrelevant elements of the initial array for given n and m. On the second line print all such i that i-th element is irrelevant. Numbers on the second line must be printed in the ascending order and must be separated by spaces.

Sample Input

3 2

Sample Output

12

Source

Northeastern Europe 2004

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唯一分解定理+组合数~

题意为一个长为n的数列，每次合并相邻两个数，合并n-1次成为一个数，问最后这个数除m的余数与哪些数无关。

所有的数中，系数是m的倍数的数即为所求，所以题目可以转化为c(n-1,1)到c(n-1,n-1)中是m的倍数的数。

把m分解，记录系数，再与各组合数的次数对比，小于该组合数时，该组合数可行。

很巧妙的一点是在分解时直接o(n)循环即可，每次除掉i，这样除出来的数一定是质数，因为合数会在循环到它的质因子时除掉。

#include<cstdio>#include<cstring>#define ll long longll n,m,a[2][100001],cnt,tmp[100001],ans[100001];int main(){scanf("%lld%lld",&n,&m);for(int i=2;i*i<=m;i++)  if(!(m%i))  {  a[0][++cnt]=i;  while(!(m%i)) m/=i,a[1][cnt]++;  }if(m!=1) a[0][++cnt]=m,a[1][cnt]=1;for(int i=1;i<n-1;i++){int x=n-i,y=i;for(int j=1;j<=cnt;j++)  if(!(x%a[0][j]))  {  while(!(x%a[0][j])) x/=a[0][j],tmp[j]++;  }for(int j=1;j<=cnt;j++)  if(!(y%a[0][j]))  {  while(!(y%a[0][j])) y/=a[0][j],tmp[j]--;  }int flag=0;for(int j=1;j<=cnt;j++)  if(a[1][j]>tmp[j])  {  flag=1;break;  }if(!flag) ans[++ans[0]]=i+1;}printf("%lld\n",ans[0]);for(int i=1;i<=ans[0];i++) printf("%lld ",ans[i]);printf("\n");return 0;}