• 模板题Pollard_Rho大数分解 A


    题意:是素数就输出Prime,不是就输出最小因子.

    #include <cstdio>
    #include<time.h>
    #include <algorithm>
    #include<set>
    using namespace std;
    
    typedef long long llt;
    
    int const Repeat = 10;
    set<llt>sss;
    //利用二进制计算a*b%mod
    llt multiMod(llt a, llt b, llt mod){
        llt ret = 0LL;
        a %= mod;
        while (b){
            if (b & 1LL) ret = (ret + a) % mod, --b;
            b >>= 1LL;
            a = (a + a) % mod;
        }
        return ret;
    }
    
    //计算a^b%mod
    llt powerMod(llt a, llt b, llt mod){
        llt ret = 1LL;
        a %= mod;
        while (b){
            if (b & 1LL) ret = multiMod(ret, a, mod), --b;
            b >>= 1LL;
            a = multiMod(a, a, mod);
        }
        return ret;
    }
    
    //Miller-Rabin测试,测试n是否为素数
    bool Miller_Rabin(llt n, int repeat){
        if (2LL == n || 3LL == n) return true;
        if (!(n & 1LL)) return false;
    
        //将n分解为2^s*d
        llt d = n - 1LL;
        int s = 0;
        while (!(d & 1LL)) ++s, d >>= 1LL;
    
        //srand((unsigned)time(0));
        for (int i = 0; i<repeat; ++i){//重复repeat次
            llt a = rand() % (n - 3) + 2;//取一个随机数,[2,n-1)
            llt x = powerMod(a, d, n);
            llt y = 0LL;
            for (int j = 0; j<s; ++j){
                y = multiMod(x, x, n);
                if (1LL == y && 1LL != x && n - 1LL != x) return false;
                x = y;
            }
            if (1LL != y) return false;
        }
        return true;
    }
    
    llt Fac[100];//质因数分解结果(刚返回时是无序的)
    int FCnt;//质因数的个数。数组小标从0开始
    
    llt gcd(llt a, llt b){
        if (0L == a || 0L == b) return 1;
        if (a < 0) a = -a;
        if (b < 0) b = -b;
        while (b){
            llt t = a % b;
            a = b;
            b = t;
        }
        return a;
    }
    llt Pollard_Rho(llt n, llt c){
        llt i = 1, k = 2;
        llt x = rand() % n;
        llt y = x;
        while (1){
            ++i;
            x = (multiMod(x, x, n) + c) % n;
            llt d = gcd(y - x, n);
            if (d != 1LL && d != n) return d;
            if (y == x) return n;
            if (i == k) y = x, k <<= 1;
        }
    }
    
    void find(llt n){
        if (4LL == n){
            Fac[0] = Fac[1] = 2LL;
            FCnt = 2;
            return;
        }
        if (Miller_Rabin(n, Repeat)){
            Fac[FCnt++] = n;
            return;
        }
    
        llt p;
        while ((p = Pollard_Rho(n, rand() % (n - 3) + 3)) == n);
    
        find(p);
        find(n / p);
    }
    
    int main(){
        int kase;
        scanf("%d", &kase);
        while (kase--){
            llt n;
            scanf("%lld", &n);
    
            FCnt = 0;
            if (Miller_Rabin(n, 10)){ printf("Prime
    "); }
            else{
                find(n);
                llt ans = Fac[0];
                for (int i = 0; i < FCnt;++i)
                if (ans>Fac[i])ans = Fac[i];
                printf("%lld
    ", ans);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ALINGMAOMAO/p/9833304.html
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