题意 给出a d n    给出数列 a,a+d,a+2d,a+3d......a+kd 问第n个数是几 保证答案不溢出 直接线性筛模拟即可 #include<cstdio> #include<cstring> using namespace std; ]; ]; ]; int cnt; void Prime(int n){ cnt=; memset(Is_Primes,,sizeof(Is_Primes)); ;i<=n;i++){ if(!Is_Primes[i]) Pr…

题意 哥德巴赫猜想:任一大于2的数都可以分为两个质数之和 给一个n 分成两个质数之和 线行筛打表即可 可以拿一个数组当桶标记一下a[i]  i这个数是不是素数  在线性筛后面加个装桶循环即可 #include<cstdio> #include<cstring> using namespace std; ]; ]; int cnt; void Prime(int n){ cnt=; memset(Is_Primes,,sizeof(Is_Primes)); ;i<=n;i++)…

题意:给一个数 可以写出多少种  连续素数的合 思路:直接线性筛 筛素数 暴力找就行   (素数到n/2就可以停下了,优化一个常数) 其中:线性筛的证明参考:https://blog.csdn.net/nk_test/article/details/46242401 https://blog.csdn.net/qq_40873884/article/details/79124552 https://blog.csdn.net/baoli1008/article/details/50788512…

题目地址:http://poj.org/problem?id=3006 刷了好多水题,来找回状态...... Dirichlet's Theorem on Arithmetic Progressions Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16803   Accepted: 8474 Description If a and d are relatively prime positive integers, t…

Dirichlet's Theorem on Arithmetic Progressions Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13636   Accepted: 6808 Description If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing…

Dirichlet's Theorem on Arithmetic Progressions Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16733   Accepted: 8427 Description If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing…

http://poj.org/problem?id=3006 #include <cstdio> using namespace std; bool pm[1000002]; bool usd[1000002]; bool judge(int x) { if(usd[x])return pm[x]; usd[x] = true; if(x == 2) return pm[x] = true; if(((x & 1) == 0) || (x < 2))return pm[x] =…

题目大意:a和d是两个互质的数,则序列a,a+d,a+2d,a+3d,a+4d ...... a+nd 中有无穷多个素数,给出a和d,找出序列中的第n个素数 #include <cstdio> int isPrime(int n) { || (n != && n % == )) ; ; i*i <= n; ++i) ) ; ; } int main() { int a,d,n,cun,temp; while(scanf("%d %d %d",&…

http://poj.org/problem?id=3006 #include<stdio.h> #include<math.h> int is_prime(int n) { int i,m; ) ; m = sqrt(n); ; i <= m; i ++) { ) ; } ; } int main() { int a,d,n,i; while(~scanf("%d%d%d",&a,&d,&n)) { ; &&d==…

简单的暴力筛法就可. #include <iostream> #include <cstring> #include <cmath> #include <cctype> #include <cstdio> #include <cmath> #include <algorithm> #include <numeric> using namespace std; , M = ; bool is[N]; int pr…