Prime Path素数筛与BFS动态规划
- 2020 年 5 月 8 日
- 筆記
- 刷题笔记: POJ, 算法----------
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。
C++实现
#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;
}
整除问题
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。
输入描述
两个整数n(2<=n<=1000)，a(2<=a<=1000)
输出描述
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
*/
拓展
Prime Path素数筛与BFS动态规划的综合应用
问题 POJ
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
*/
