Dirichlet's Theorem on arithmetic sions ssions
| Time limit:1000 ms |
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Memory limit:65536 K |
| Total submissions:15714 |
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Accepted:7887 |
Description
IfAAndDAre relatively prime positive integers, the arithmetic sequence beginningAAnd increasingD, I. e .,A,A+D,A+ 2D,A+ 3D,A+ 4D,..., Contains infinitely implements prime numbers. this fact is known as Dirichlet's Theorem on arithmetic progressions, which had been conjectured by Johann Carl Friedrich Gauss (1777-1855) and was proved by Johann Peter Gustav Lejeune Dirichlet (1805-1859) in 1837.
For example, the arithmetic sequence beginning with 2 and increasing by 3, I. e .,
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98 ,...,
Contains infinitely invalid prime numbers
2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89 ,....
Your mission, shocould you decide to accept it, is to write a program to findNTh prime number in this arithmetic sequence for given positive integersA,D, AndN.
Input
The input is a sequence of datasets. a dataset is a line containing three positive integersA,D, AndNSeparated by a space.AAndDAre relatively prime. You may assumeA<= 9307,D<= 346, andN<= 210.
The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.
Output
The output shoshould be composed of as many lines as the number of the input datasets. Each line shoshould contain a single integer and shoshould never contain extra characters.
The output integer corresponding to a datasetA,D,NShocould beNTh prime number among those contained in the arithmetic sequence beginningAAnd increasingD.
FYI, it is known that the result is always less than 106 (one million) under this input condition.
Sample Input
367 186 151179 10 203271 37 39103 230 127 104 185253 50 851 1 19075 337 210307 24 79331 221 177259 170 40269 58 1020 0 0
Sample output
928096709120371039352314503289942951074127172269925673
Source
Japan 2006 domestic
Question:
Input three numbers a, B, N. To obtain the prime number in the nth sequence, that is, the prime number sequence in the deviation series with a as the first item and B as the tolerances.
#include <iostream>#include <math.h>#include <string>using namespace std;int main(){int a,d,n;while(cin>>a>>d>>n&&(a||d||n)){int i,j,k=1,sushu[100001];int m;for(i=a;i<=1000001;i=i+d){m=int(sqrt((double)i)); for(j=2;j<=m;j++){if(i%j==0)break;}if(j>m&&i!=1){sushu[k]=i;if(k==n)break;k++;}}cout<<sushu[n]<<endl;}return 0;}