POJ 3641 Pseudoprime numbers
来源:互联网 发布:福建广电网络人工电话 编辑:程序博客网 时间:2024/03/29 13:36
Pseudoprime numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7396 Accepted: 3050
Description
Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)
Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.
Input
Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.
Output
For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".
Sample Input
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
Sample Output
no
no
yes
no
yes
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7396 Accepted: 3050
Description
Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)
Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.
Input
Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.
Output
For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".
Sample Input
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0
Sample Output
no
no
yes
no
yes
yes
数论水题。如果p是素数,不是伪素数,否则,如果满足a的p次方取余p等于a,则p是伪素数,剩下的情况均不是。
#include <iostream>#include <cstdio>using namespace std;bool is_prime(long long int m){ for(long long int i=2;i*i<=m;i++) if(m%i==0) return false; return true;}long long int mod_pow(long long int x,long long int n,long long int mod){long long int res=1;while(n>0){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}int main(void){ long long int p,a; while(scanf("%lld%lld",&p,&a)!=EOF&&(p||a)) { if(is_prime(p)) printf("no\n"); else{ if(a==mod_pow(a,p,p)) printf("yes\n"); else printf("no\n");} }}
0 0
- POJ-3641-Pseudoprime numbers
- POJ 3641-Pseudoprime numbers
- poj 3641 Pseudoprime numbers
- poj 3641 Pseudoprime numbers
- POJ 3641 Pseudoprime numbers
- poj 3641 Pseudoprime numbers
- POJ-3641 Pseudoprime numbers
- POJ 3641 Pseudoprime numbers
- 【POJ 3641】Pseudoprime numbers
- poj 3641 Pseudoprime numbers
- POJ-3641 Pseudoprime numbers
- Pseudoprime numbers POJ 3641
- poj 3641 Pseudoprime numbers
- poj 3641 Pseudoprime numbers
- POJ - 3641 Pseudoprime numbers
- 【 POJ 3641 】Pseudoprime numbers
- 【POJ 3641】Pseudoprime numbers
- 【POJ】[3641]Pseudoprime numbers
- Java回顾之I/O
- ListView上拉加载更多
- 批量清除WPS/Word文档中的回车符
- jQuery学习之AJAX
- 第三周工作周报
- POJ 3641 Pseudoprime numbers
- LeetCode(20)Valid Parentheses
- GC垃圾收集算法
- Grep用法
- hdu 3593
- C++ class与内存
- LeetCode(141)(142) Linked List Cycle I II
- Linux - 有效群组(effective group)与初始群组(initial group),groups,newgrp
- 高仿SinaWeibo新浪微博发布页面话题效果