标签:分解 uva close namespace 计算 play prim one name
两个超大组合数相除。解法是分解质因数,用数组记录每个质因数称或除的次数,最后直接遍历数组计算即可
#include<iostream> #include<cstdio> #include<cmath> #include<vector> using namespace std; void Fact(int n, int opt, vector<int> primes, vector<int>& fact); void addFact(int n, int opt, vector<int> primes, vector<int>& fact); int main() { int p, q, r, s; while (cin >> p >> q >> r >> s) { vector<int> nums(10001, 1); vector<int> primes(1); //获取1-10000的所有质数 for (int i = 2; i <= 10000; ++i) { if (nums[i] == 1) { primes.push_back(i); for (int i1 = 1; i*i1 < 10000; ++i1) { nums[i*i1] = 0; } } } vector<int> fact(primes.size(), 0); addFact(p, 1, primes, fact); addFact(q, -1, primes, fact); addFact(p - q, -1, primes, fact); addFact(r, -1, primes, fact); addFact(s, 1, primes, fact); addFact(r - s, 1, primes, fact); double answer = 1; for (int i = 1; i < fact.size(); ++i) { if (fact[i] != 0) { answer *= pow(primes[i], fact[i]); } } printf("%.5lf\n", answer); } return 0; } void Fact(int n,int opt,vector<int> primes,vector<int>& fact) { for (int i = 1; i < primes.size(); ++i) { while (!(n%primes[i])) { n = n / primes[i]; fact[i]+=opt; } if (n == 1) { break; } } } void addFact(int n, int opt, vector<int> primes, vector<int>& fact) { for (int i = 1; i <= n; ++i) { Fact(i, opt, primes, fact); } }
(用JAVA的大数好像直接算就可以而且不会超时)
标签:分解 uva close namespace 计算 play prim one name
原文地址:http://www.cnblogs.com/Algorithm-X/p/7522172.html