• HDU-5778 abs


    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;
    }


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  • 原文地址:https://www.cnblogs.com/wiklvrain/p/8179431.html
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