【POJ3006】Dirichlet's Theorem on Arithmetic Progressions（素數篩法）

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簡單的暴力篩法就可。

 1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #include <cctype> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <numeric> 9 using namespace std;10 11 const int N = 1000005, M = 500005;12 bool is[N]; int prm[M];13 int getprm (int n) {14     int i, j, k = 0;15     int s, e = (int)(sqrt (0.0 + n) + 1);16     memset (is, 1, sizeof(is));17     prm[k ++] = 2; is[0] = is[1] = 0;18     for (i = 4; i < n; i += 2) is[i] = 0;19     for (i = 3; i < e; i += 2) {20         if (is[i]) {21             prm[k ++] = i;22             for (s = i * 2, j = i * i; j < n; j += s) {23                 is[j] = 0;24             }25         }26     }27     for (; i < n; i += 2) {28         if (is[i]) prm[k ++] = i;29     }30     return k;31 }32 33 int main () {34     ios :: sync_with_stdio(false);35     //freopen("out.txt", "w", stdout);36     getprm(1000000);37     /*for (int i = 0 ; i < 78499; ++ i) {38         cout << i  << " : " << prm[i] << endl;39     }*/40     long long a, d, n;41     long long cnt = 0, cur = 0;42     while (cin >> a >> d >> n) {43         if (a == 0 && d == 0 && n == 0) break;44         cnt = 0 ;45         while (1) {46             if (binary_search(prm, prm + 78499, a)) {47                 cnt ++;48                 if (cnt == n) break;49             }50             a += d;51             //cout << "step: " << cnt << " " << a << endl;52         }53         cout << a << endl;54     }55     return 0;56 }

 

【POJ3006】Dirichlet's Theorem on Arithmetic Progressions（素數篩法）