题目大意:
给定一个数x,求正整数ygeq 2y≥2,使得满足以下条件:
1.y-x的绝对值最小
2.y的质因数分解式中每个质因数均恰好出现2次。
解题思路:
由于y质因数分解式中每个质因数均出现2次,那么y是一个完全平方数,设y=z*z,题目可转换成求z,使得每个质因数出现1次. 我们可以暴力枚举z,检查z是否符合要求,显然当z是质数是符合要求,由素数定理可以得,z的枚举量在logn级别
代码:
朴素写法:
#include <cmath> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long LL; const int MAXN = 1e5 + 10; LL prime[MAXN]; void init() { memset(prime, 0, sizeof(prime)); for(int i = 2; i < MAXN; ++i){ if(!prime[i]) prime[++prime[0]] = i; for(int j = 1; j <= prime[0] && prime[j] < MAXN / i; ++j){ prime[prime[j] * i] = 1; if(i % prime[j] == 0) break; } } } bool isPrime(LL n){ for(int i = 2; i * i <= n; ++i){ if(n % i == 0) return 0; } return 1; } bool Judge(LL n){ for(int i = 1; i <= prime[0]; ++i){ if(n % prime[i] == 0){ n /= prime[i]; if(n % prime[i] == 0) return 0; } } if(n == 1 || isPrime(n)) return 1; else return 0; } LL solve(LL n){ LL d = sqrt(n + 0.5); LL ans = 1e18; for(LL i = d; i >= 2; --i){ if(Judge(i)){ ans = min(ans, n - (i * i)); break; } } for(LL i = d + 1; ; ++i){ if(Judge(i)){ ans = min(ans, (i * i) - n); break; } } return ans; } int main(){ LL n, t; cin >> t; init(); while(t--){ cin >> n; cout << solve(n) << endl; } return 0; }
Miller_Rabin + Pollard_rho写法:
#include <cmath> #include <time.h> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef long long LL; const int maxn = 100 + 5; const int S = 20; int tot; LL factor[maxn]; LL mod_mul(LL a, LL b, LL n){ LL res = 0; while(b){ if(b & 1) res = (res + a) % n; a = (a + a) % n; b >>= 1; } return res; } LL mod_exp(LL a, LL b, LL n){ LL res = 1; while(b){ if(b & 1) res = mod_mul(res, a, n); a = mod_mul(a, a, n); b >>= 1; } return res; } LL gcd(LL a, LL b){ if(a == 0) return 1; if(a < 0) return gcd(-a, b); while(b){ LL tmp = a % b; a = b; b = tmp; } return a; } LL pollard_rho(LL x, LL c){ LL i = 1, k = 2; LL x0 = rand() % x; LL y = x0; while(true){ ++i; x0 = (mod_mul(x0, x0, x) + c) % x; LL d = gcd(y - x0, x); if(d != 1 && d != x) return d; if(y == x0) return x; if(i == k){ y = x0; k += k; } } } bool miller_rabin(LL n) { if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11) return true; if(n == 1 || !(n % 2) || !(n % 3) || !(n % 5) || !(n % 7) || !(n % 11)) return false; LL x, pre, u = n - 1, k = 0; while(!(u & 1)){ ++k; u >>= 1; } srand((LL)time(NULL)); for(int i = 0; i < S; ++i){ //进行S次测试,S越大,结果越准确 x = rand() % (n - 2) + 2; //在[2, n)中取随机数 if(x % n == 0) continue; x = mod_exp(x, u, n); //计算x^u % n pre = x; for(int j = 0; j < k; ++j){ x = mod_mul(x, x, n); if(x == 1 && pre != 1 && pre != n - 1) return false; pre = x; } if(x != 1) return false; } return true; } void findfactor(LL n){ if(miller_rabin(n)){ factor[tot++] = n; return; } LL p = n; while(p >= n){ p = pollard_rho(p, rand() % (n - 1) + 1); } findfactor(p); findfactor(n / p); } int main(){ LL t, x, tmp; cin >> t; while(t--){ cin >> x; if(x == 1 || x == 2 || x == 3 || x == 4){ cout << (4 - x) << endl; continue; } tmp = (LL)sqrt(x + 0.500); LL ans = 1e18; for(LL i = tmp; i > 1; --i){ tot = 0; memset(factor, 0, sizeof(factor)); findfactor(i); sort(factor, factor + tot); int flag = 1; for(int j = 1; j < tot; ++j){ if(factor[j] == factor[j-1]) {flag = 0; break;} } if(flag) {ans = min(ans, x - i * i); break;} } for(LL i = tmp + 1; ; ++i){ tot = 0; memset(factor, 0, sizeof(factor)); findfactor(i); sort(factor, factor + tot); int flag = 1; for(int j = 1; j < tot; ++j){ if(factor[j] == factor[j-1]) {flag = 0; break;} } if(flag) {ans = min(ans, i * i - x); break;} } cout << ans << endl; } return 0; }