埃拉托斯特尼筛法（sieve of Eratosthenes ） 是古希腊数学家埃拉托斯特尼发明的计算素数的方法。对于求解不大于n的所有素数，我们先找出sqrt(n)内的所有素数p1到pk，其中k = sqrt(n)，依次剔除Pi的倍数，剩下的所有数都是素数。
具体操作如上述 图片所示。
#include<iostream>
#include<vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<bool> isprime(n + 5, true);
vector<int> ans;
for (int i = 2; i <= n; i++) {
if (isprime[i]) {
ans.push_back(i);
for (int j = i * i; j <= n; j += i)isprime[j] = false;
}
}
for (auto i : ans)cout << i << " ";
cout << endl;
return 0;
}
给定n，a求最大的k，使n！可以被ak整除但不能被a(k+1)整除。
两个整数n(2<=n<=1000)，a(2<=a<=1000)
示例1
输入
555 12
输出
274
#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main() {
int n, a, temp;
int ans = 0x7fffffff;
cin >> n >> a;
vector<bool> isprime(1010, true);
vector<int> prime; //素数列表
map<int, int> primecntnp; //存储n!的质因子的指数
map<int, int> primecnta; //存储a的质因子的指数
for (int i = 2; i <= 1010; i++) { //采用素数筛选出前1010个数中的素数，并将map初始化
if (isprime[i]) {
prime.push_back(i);
primecntnp[i] = primecnta[i] = 0;
for (int j = i * i; j <= 1010; j += i)isprime[j] = false;
}
}
//4! = 24 = 1*2*3*4 = 2*2*2*3
for (int i = 0; i < prime.size(); i++) { //对n!进行因式分解
temp = n;
while (temp) { //按照p、p*p、p*p*p来进行因式分解
primecntnp[prime[i]] += temp / prime[i];
temp /= prime[i];
}
}
for (int i = 0; i < prime.size(); i++) { //对a进行因式分解
temp = a;
while (temp % prime[i] == 0) {
primecnta[prime[i]]++;
temp /= prime[i];
}
if (primecnta[prime[i]] == 0)continue; //a里面不存在的则无法提供
if (primecntnp[prime[i]] / primecnta[prime[i]] < ans)ans = primecntnp[prime[i]] / primecnta[prime[i]];
}//找到最小的指数，便是最大的k值
cout << ans << endl;
return 0;
}
/*
555 12
274
*/
Prime Path素数筛与BFS动态规划的综合应用
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on… Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
问题大意
从一个素数换到另一个素数，每次只能换一个数字（一位）且换后的每次都是素数。求最小次数？
C++代码
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 10000;
bool isprime[maxn + 1];
int dp[maxn + 1];
int getNext(int num, int t, int change){
//num : 当前的数，t当前的位置，change是改变位的值
if(t == 0) return num / 10 * 10 + change; //最低位
else if(t == 1) return num /100 * 100 + change * 10 + num % 10;
else if(t == 2) return num /1000 * 1000 + change * 100 + num % 100;
else return change * 1000 + num % 1000;
}
int main(){
fill(isprime+2, isprime + maxn, true);
for(int i = 2; i <= maxn; i++){
if(isprime[i]){
for(int j = i * i; j <= maxn; j += i){
isprime[j] = false;
}
}
}//打表
int T;
cin>>T;
while(T--){
int a, b;
cin>>a>>b;
fill(dp, dp + maxn, 0x3f);
dp[a] = 0; //记录从一个prime跳跃到另一个prime所需的最少次数
queue<int> q;
q.push(a);
while(!q.empty()){
int cur = q.front(); //取出队列的第一个
q.pop();
for(int i = 0; i < 4; i++){
for(int j = 0; j < 10; j++){
if(i == 3 && j == 0) continue; //
int next = getNext(cur, i, j); //替换
if(isprime[next] == false || dp[next] <= dp[cur]) continue;
// 不是素数不行，如果到next已经有更小的那也不用这个变换路径了
dp[next] = dp[cur] + 1;
q.push(next);
}
}
}
cout<<dp[b]<<endl;
}
return 0;
}
/*
3
1033 8179
1373 8017
1033 1033
*/
原文：https://www.cnblogs.com/RioTian/p/12853123.html