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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 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 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 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. 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 Sample Input 3 2 Sample Output 1 2 Source |
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=100100;
int n,m;
int pm[maxn],pr[maxn],pc[maxn],pn;
vector<int> ans;
void f(int x,int d)
{
for(int i=0;i<pn;i++)
while(x%pm[i]==0)
{
pc[i]+=d;
x/=pm[i];
}
}
bool check()
{
for(int i=0;i<pn;i++)
{
if(pc[i]<pr[i]) return false;
}
return true;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
ans.clear(); pn=0;
/**********/
for(int i=2;i*i<=m;i++)
{
if(m%i==0)
{
pm[pn]=i;
while(m%i==0)
{
pr[pn]++;
m/=i;
}
pn++;
}
}
if(m!=1)
{
pm[pn]=m;
pr[pn++]=1;
}
/**********/
n--;
for(int i=1;i<=n;i++)
{
f(n-i+1,1);
f(i,-1);
if(check())
{
ans.push_back(i+1);
}
}
printf("%d\n",ans.size());
sort(ans.begin(),ans.end());
for(int i=0,sz=ans.size();i<sz;i++)
{
printf("%d ",ans[i]);
}
putchar(10);
}
return 0;
}