題解：http://vjudge.net/problem/viewProblem.action?id=43952

把<10^6的素數先搞出來，然后枚舉每個

對于p
篩掉(a * n + b) % p = 0的
分兩種情況
1. a % p = 0 的話，(a * n + b) % p = b % p，
? ?b % p = 0 的話才能使 (a * n + b) % p = 0 ，然后直接答案為0
2. a % p != 0 的話，先求出最小的n0，使得(a * n0 + b) % p = 0
? ? 然后(a * (n0 + x) + b) % p = (a * x) % p = 0 ，p是素數，所以x 是p的倍數。。。
? ? 然后篩掉n0, n0 + p, n0 + 2 * p....

最后統計下就行了

#include <iostream> #include <vector> #include <algorithm>using namespace std;// maximum prime trial divisor we need (up to sqrt of max a*U+b) const long long MAX_PRIME = 1000000;// maximum range of values of n to consider (U-L+1) const long long MAX_RANGE = 1000000+1;vector<int> primes; bool prime[MAX_PRIME+1];long long gcd(long long a, long long b) {if (!b) return a;return gcd(b, a%b); }long long gcd(long long a, long long b, long long &s, long long &t) {long long r, r1, r2, a1, a2, b1, b2, q;a1 = b2 = 1;a2 = b1 = 0;while (b) {q = a / b;r = a % b;r1 = a1 - q*b1;r2 = a2 - q*b2;a = b;a1 = b1;a2 = b2;b = r;b1 = r1;b2 = r2;}s = a1;t = a2;return a; }// generate all primes up to max(sqrt(a*U+b)) void genprimes() {fill(prime, prime+MAX_PRIME+1, true);prime[0] = prime[1] = false;// use a sievefor (long long p = 2; p < MAX_PRIME+1; p++) {if (!prime[p]) continue;primes.push_back(p);for (long long k = p*p; k < MAX_PRIME+1; k += p) {prime[k] = false;}}//cout<<primes.size()<<endl; }bool isprime(long long n) {// if it's below max prime, just check the sieveif (n <= MAX_PRIME) {return prime[n];}// trial division for the restfor (unsigned int i = 0; i < primes.size(); i++) {if (n % primes[i] == 0) return false;}return true; }bool solve(int id) {long long a, b, L, U;cin >> a;if (!a) return false;cin >> b >> L >> U;cout << "Case " << id << ": ";// special case: if gcd(a, b) > 1, then the only potential prime is when// n == 0 and b is prime.if (gcd(a, b) > 1) {if (L == 0 && isprime(b)) {cout << 1 << endl;} else {cout << 0 << endl;}return true;}// gcd(a, b) = 1, do a sieve on range of nbool nprime[MAX_RANGE];int range = U - L + 1;fill(nprime, nprime + range, true);// special cases//// if a*n+b <= 2 we will have to account for that//// n = 0 (only occurs if L = 0)// n = 1 (if L <= 1)if (L == 0) {nprime[0] = isprime(b);}if (L <= 1) {nprime[1-L] = isprime(a+b);}long long maxval = a * U + b;for (unsigned int i = 0; i < primes.size(); i++) {long long p = primes[i];// p cannot divide b since gcd(a,b) = 1if (a % p == 0) continue;// p cannot divide more a*n+bif (p*p > maxval) break;// solve for n in a*n+b = 0 mod plong long s, t;gcd(a, p, s, t);int n_start = (-(b*s)) %p;if (n_start < 0) n_start += p;int Lrem = L % p;n_start -= Lrem;if (n_start < 0) n_start += p;// get past the first multiple which is prime!while (a*(n_start + L)+b <= p) {n_start += p;}// finally do the sievewhile (n_start < range) {nprime[n_start] = false;n_start += p;}}#ifdef DEBUGfor (int i = 0; i < range; i++) {if (nprime[i]) {cout << " " << a*(i+L)+b << endl;}} #endifcout << count(nprime, nprime+range, true) << endl;return true; }int main() {genprimes();int id = 1;while (solve(id++));return 0; }

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