Dirichlet's Theorem on Arithmetic Progressions
2018-01-20 11:56
357 查看
a ,d ,n给定质数a,d
求数列a,a+d,a+2d,.....的第N个素数(不超过1000000)
筛法,然后遍历一边即可,不知道单纯模拟会不会超
#include<stdio.h>
#include<string.h>
int p[1000001];
int main()
{
int i,c,a,d,n,j;
memset(p,0,sizeof(p));p[0]=1;p[1]=1;
for (i=2;i<=1000001;i++)
{
if (p[i]==0)
for (j=2*i;j<1000001;j+=i)
p[j]=1;
}
while (1)
{
scanf("%d%d%d",&a,&d,&n);
if (a+d+n==0) break;
for (i=a,c=0;c<n;i+=d)
if (!p[i]) {c++;if (c==n) printf("%d\n",i);}
}
}
求数列a,a+d,a+2d,.....的第N个素数(不超过1000000)
筛法,然后遍历一边即可,不知道单纯模拟会不会超
#include<stdio.h>
#include<string.h>
int p[1000001];
int main()
{
int i,c,a,d,n,j;
memset(p,0,sizeof(p));p[0]=1;p[1]=1;
for (i=2;i<=1000001;i++)
{
if (p[i]==0)
for (j=2*i;j<1000001;j+=i)
p[j]=1;
}
while (1)
{
scanf("%d%d%d",&a,&d,&n);
if (a+d+n==0) break;
for (i=a,c=0;c<n;i+=d)
if (!p[i]) {c++;if (c==n) printf("%d\n",i);}
}
}
相关文章推荐
- POJ3006 Dirichlet's Theorem on Arithmetic Progressions【筛选法】
- POJ 3006 Dirichlet's Theorem on Arithmetic Progressions
- POJ 3006 Dirichlet's Theorem on Arithmetic Progressions
- POJ 刷题系列:3006. Dirichlet's Theorem on Arithmetic Progressions
- *寒假水97——Dirichlet's Theorem on Arithmetic Progressions
- POJ 3006 Dirichlet's Theorem on Arithmetic Progressions
- O - Dirichlet's Theorem on Arithmetic Progressions
- O - Dirichlet's Theorem on Arithmetic Progressions
- 《数据结构编程实验》 1.5.5Dirichlet's Theorem on Arithmetic Progressions
- (素数求解)I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)
- 3006. Dirichlet's Theorem on Arithmetic Progressions
- POJ3006_Dirichlet's Theorem on Arithmetic Progressions_筛法求素数表
- O - Dirichlet's Theorem on Arithmetic Progressions
- Dirichlet's Theorem on Arithmetic Progressions
- Dirichlet's Theorem on Arithmetic Progressions
- POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数的判断 筛选法
- POJ 3006 - Dirichlet's Theorem on Arithmetic Progressions
- I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)
- poj 3006 Dirichlet's Theorem on Arithmetic Progressions
- Dirichlet's Theorem on Arithmetic Progressions 筛取素数