标签:io ar for sp on html amp ad ef
输入a b 求有多少对p, q 使得p*q == a && p < q && p >= b
直接大整数分解 然后dfs搜出所有可能的解
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <cmath> using namespace std; typedef long long LL; const int Times = 25; LL factor[100], f[100]; int l, ll, ans, num[100]; LL a, b; LL gcd(LL a, LL b) { return b ? gcd(b, a%b):a; } LL add_mod(LL a, LL b, LL n) { LL ans = 0; while(b) { if(b&1) ans = (ans + a)%n; b >>= 1; a = (a<<1)%n; } return ans; } LL pow_mod(LL a, LL m, LL n) { LL ans = 1; while(m) { if(m&1) ans = add_mod(ans, a, n); m >>= 1; a = add_mod(a, a, n); } return ans; } bool Witness(LL a, LL n) { int j = 0; LL m = n-1; while(!(m&1)) { j++; m >>= 1; } LL x = pow_mod(a, m, n); if(x == 1 || x == n-1) return true; while(j--) { x = add_mod(x, x, n); if(x == n-1) return true; } return false; } bool Miller_Rabin(LL n) { if(n < 2) return false; if(n == 2) return true; if(!(n&1)) return false; for(int i = 0; i < Times; i++) { LL a = rand()%(n-1)+1; if(!Witness(a, n)) return false; } return true; } LL Pollard_rho(LL n, LL c) { LL i = 1, x = rand()%(n-1)+1, y = x, k = 2, d; //srand(time(NULL)); while(true) { i++; x = (add_mod(x,x,n)+c)%n; d = gcd(y-x,n); if(d > 1 && d < n) return d; if(y == x) return n; if(i == k) { y = x; k <<= 1; } } } void get_fact(LL n, LL k) { if(n == 1) return; if(Miller_Rabin(n)) { factor[l++] = n; return; } LL p = n; while(p >= n) { p = Pollard_rho(p, k--); } get_fact(p, k); get_fact(n/p, k); } void dfs(LL x, int p, LL m) { if(x > m) return; if(p == ll) { if(x >= b && a/x > x) ans++; //printf("%lld\n", x); return; } LL y = 1; for(int i = 0; i <= num[p]; i++) { dfs(x*y, p+1, m); y *= f[p]; } } int main() { int cas = 1; int T; scanf("%d", &T); while(T--) { scanf("%lld %lld", &a, &b); LL m = sqrt(a+0.5); ans = 0; l = 0; get_fact(a, 120); sort(factor, factor+l); f[0] = factor[0]; num[0] = 1; ll = 1; for(int i = 1; i < l; i++) { if(factor[i] != factor[i-1]) { ll++; f[ll-1] = factor[i]; num[ll-1] = 0; } num[ll-1]++; } //for(int i = 0; i < ll; i++) // printf("%lld %d\n", f[i], num[i]); dfs(1, 0, m); printf("Case %d: %d\n", cas++, ans); } return 0; }
Light OJ 1341 Aladdin and the Flying Carpet Pollard_rho整数分解+DFS
标签:io ar for sp on html amp ad ef
原文地址:http://blog.csdn.net/u011686226/article/details/39931205