UVa1635 - Irrelevant Elements（质因数分解）

最新推荐文章于 2024-08-02 19:51:34 发布

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最新推荐文章于 2024-08-02 19:51:34 发布

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分类专栏：数论质因数分解组合数学

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本文探讨了一种由Georgie提出的新型随机数生成方案，该方案通过一系列操作最终得到一个介于0到m-1之间的随机数。文章重点在于如何确定初始数组中哪些元素对于最终结果是无关紧要的，并提供了一个具体的算法实现。

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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,..., a_n. 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} + a_n. 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 a_i. He considers various n and m and wonders which elements are irrelevant for these parameters. Help him to find it out.

Input

Input file contains several datasets. Each datasets has n and m ( 1 $\le$ n $\le$ 100 000, 2 $\le$ m $\le$ 10⁹) in a single line.

Output

On the first line of the output for each dataset print the number of irrelevant elements of the initial array for given n and m. On the second line print all such ithat 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

1
2

题意：整个式子的和可以化简为 sigma (C(n-1,i-1)*ai)

思路：只要判断C(n-1,i-1)能否被 m整除即可。

做法是先分解m的质因数，然后计算1！~(n-1)! 包含m的质因数的个数

C（n-1,i-1） = (n-1)!/((i-1)!*(n-i)!)

只要判断剩下的质因数的个数是否大于等于m的任一个质因数的个数即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxp = 40000+10;
const int maxn = 100000+10;
int n,m;
bool isPrime[maxp];
vector<int> ret,prime,hp,cnt,tmp;
vector<int> g[maxn];
void getPrime(){
    memset(isPrime,true,sizeof isPrime);
    for(int i = 2; i < maxp; i++){
        if(isPrime[i]){
            prime.push_back(i);
            for(int j = i+i;j < maxp; j += i){
                isPrime[j] = false;
            }
        }
    }
}
void getDigit(){
    int tk = m;
    for(int i = 0; i < prime.size() && prime[i]*prime[i] <= tk; i++){
        if(tk%prime[i]==0){
            int k = 0;
            hp.push_back(prime[i]);
            while(tk%prime[i]==0){
                tk /= prime[i];
                k++;
            }
            cnt.push_back(k);
        }
    }
    if(tk>1){
        hp.push_back(tk);
        cnt.push_back(1);
    }
}
void init(){
    cnt.clear();
    hp.clear();
    ret.clear();
    getDigit();
    for(int i = 0; i <= n-1; i++) g[i].clear();
    for(int i = 0; i <= n-1; i++) {
        for(int j = 0; j < hp.size(); j++){
            int d = 0,t = i;
            while(t) {
                d += t/hp[j];
                t /= hp[j];
            }
            g[i].push_back(d);
        }
    }
}
void solve(){
	bool miden = false;
    for(int i = 2; i <= (n-1)/2+1; i++){
        bool flag = true;
        for(int j = 0; j < hp.size(); j++){
            int d = g[n-1][j]-g[i-1][j]-g[n-i][j];
            if(d < cnt[j]){
                flag = false;
                break;
            }
        }
        if(flag) {
			ret.push_back(i);
			if(i==(n-1)/2+1 && (n&1)) miden = true;
        }
    }
    tmp.clear();
    tmp = ret;
    if(n&1){
    	int i;
    	if(miden) i = tmp.size()-2;
    	else i = tmp.size()-1;
        for(; i >= 0; i--){
            ret.push_back(n+1-tmp[i]);
        }
    }else{
        for(int i = tmp.size()-1; i >= 0; i--){
            ret.push_back(n+1-tmp[i]);
        }
    }
    printf("%d\n",ret.size());
    if(ret.size()){
        printf("%d",ret[0]);
        for(int i = 1; i < ret.size(); i++){
            printf(" %d",ret[i]);
        }

    }
    puts("");
}
int main(){
    //freopen("test.txt","r",stdin);
    getPrime();
    while(~scanf("%d%d",&n,&m)){
        init();
        solve();
    }
    return 0;
}