習題6.8 整除問題(分解質因數)

 

很好的一道題。由於n最大1000，1000的階乘過於大，故直接計算出來再判斷整除次數肯定是不行。

於是就想到了把n和a都分解質因數，冪次分別保留在不同數組中，針對a的質因數找出n和a這些質因數的冪次，看整體上n的冪次爲a的冪次的多少倍。具體來說這些質因數中n的冪次都是a的冪次的倍數，找出最小的倍數即可滿足。

 

 

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <stack>
#include <cctype>
#include <cmath>
#include <climits>
 
using namespace std;
 
const int MAXN = 1005;
const int INF = INT_MAX;

bool isPrime[MAXN];
vector<int> prime;
int prime1num[MAXN];
int prime2num[MAXN];

void Initial(){
	for(int i = 2; i < MAXN; i++){
		isPrime[i] = true;
	}
	for(int i = 2; i < MAXN; i++){
		if(!isPrime[i]) continue;
		prime.push_back(i);
		for(int j = i*i; j < MAXN; j += i){
			if(isPrime[j]) isPrime[j] = false;
		}
	}
}

int main(){
  //  freopen("in.txt", "r", stdin);
    Initial();
    int n, a, k, factor;
    while(~scanf("%d %d", &n, &a)){
    	memset(prime1num, 0, sizeof(prime1num));
    	memset(prime2num, 0, sizeof(prime2num));
    	while(n != 1){
    		int tmp = n;
	    	for(int i = 0; i < prime.size() && prime[i] <= tmp; i++){
	    		factor = prime[i];
	    		while(tmp % factor == 0){
	    			prime1num[prime[i]]++;
	    			tmp /= factor;
	    		}
	    	}
	    	n--;
	    }
	    vector<int> target;
	    for(int i = 0; i < prime.size() && prime[i] <= a; i++){
	    	factor = prime[i];
	    	bool flag = false;
	    	while(a % factor == 0){
	    		flag = true;
	    		prime2num[prime[i]]++;
	    		a /= factor;
	    	}
	    	if(flag) target.push_back(prime[i]);
	    }
	    int ans = INF;
	    for(int i = 0; i < target.size(); i++){
	    	ans = min(ans, prime1num[target[i]]/prime2num[target[i]]);
	    	//printf("%d : %d %d\n", target[i], prime1num[target[i]], prime2num[target[i]]);
	    }
	    printf("%d\n", ans);
    }
    return 0;
}