51nod1186(Miller-Rabin)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1186

题意：中文题目诶～

思路：miller_rabin模板 (详情可参见: http://blog.csdn.net/s031302306/article/details/51606018)

注意这里的 n 为 1e30，需要用字符串处理

代码：

 #include <string.h>
 #include <cstdlib>

 #define MAXL 4
 #define M10 1000000000
 #define Z10 9

 const int zero[MAXL - ] = {};

 struct bnum{
     int data[MAXL]; //  断成每截9个长度
     //  读取字符串并转存
     void read(){
         memset(data, , sizeof(data));
         char buf[];
         scanf("%s", buf);
         int len = (int)strlen(buf);
         int i = , k;
         while (len >= Z10){
             for (k = len - Z10; k < len; ++k){
                 data[i] = data[i] *  + buf[k] - '';
             }
             ++i;
             len -= Z10;
         }
         if (len > ){
             for (k = ; k < len; ++k){
                 data[i] = data[i] *  + buf[k] - '';
             }
         }
     }

     bool operator == (const bnum &x){
         return memcmp(data, x.data, sizeof(data)) == ;
     }

     bnum & operator = (const int x){
         memset(data, , sizeof(data));
         data[] = x;
         return *this;
     }

     bnum operator + (const bnum &x){
         int i, carry = ;
         bnum ans;
         for (i = ; i < MAXL; ++i){
             ans.data[i] = data[i] + x.data[i] + carry;
             carry = ans.data[i] / M10;
             ans.data[i] %= M10;
         }
         return  ans;
     }

     bnum operator - (const bnum &x){
         int i, carry = ;
         bnum ans;
         for (i = ; i < MAXL; ++i){
             ans.data[i] = data[i] - x.data[i] - carry;
             if (ans.data[i] < ){
                 ans.data[i] += M10;
                 carry = ;
             }else{
                 carry = ;
             }
         }
         return ans;
     }

     //  assume *this < x * 2
     bnum operator % (const bnum &x){
         int i;
         for (i = MAXL - ; i >= ; --i){
             if (data[i] < x.data[i]){
                 return *this;
             }
             else if (data[i] > x.data[i]){
                 break;
             }
         }
         return ((*this) - x);
     }

     bnum & div2(){
         int  i, carry = , tmp;
         for (i = MAXL - ; i >= ; --i){
             tmp = data[i] & ;
             data[i] = (data[i] + carry) >> ;
             carry = tmp * M10;
         }
         return *this;
     }

     bool is_odd(){
         return (data[] & ) == ;
     }

     bool is_zero(){
         for (int i = ; i < MAXL; ++i){
             if (data[i]){
                 return false;
             }
         }
         return true;
     }
 };

 void mulmod(bnum &a0, bnum &b0, bnum &p, bnum &ans){//计算a0*b0%p
     ans = ;
     bnum tmp = a0, b = b0;
     while (!b.is_zero()){
         if (b.is_odd()){
             ans = (ans + tmp) % p;
         }
         tmp = (tmp + tmp) % p;
         b.div2();
     }
 }

 void powmod(bnum &a0, bnum &b0, bnum &p, bnum &ans){//计算a0^b0%p
     bnum tmp = a0, b = b0;
     ans = ;
     while (!b.is_zero()){
         if (b.is_odd()){
             mulmod(ans, tmp, p, ans);
         }
         mulmod(tmp, tmp, p, tmp);
         b.div2();
     }
 }

 bool MillerRabinTest(bnum &p, int iter){
     int i, small = , j, d = ;
     for (i = ; i < MAXL; ++i){
         if (p.data[i]){
             break;
         }
     }
     if (i == MAXL){
         // small integer test
         if (p.data[] < ){
             return  false;
         }
         if (p.data[] == ){
             return  true;
         }
         small = ;
     }
     if (!p.is_odd()){
         return false;   //  even number
     }
     bnum a, s, m, one, pd1;
     one = ;
     s = pd1 = p - one;
     while (!s.is_odd()){
         s.div2();
         ++d;
     }

     for (i = ; i < iter; ++i){
         a = rand();
         if (small){
             a.data[] = a.data[] % (p.data[] - ) + ;
         }else{
             a.data[] = a.data[] / M10;
             a.data[] %= M10;
         }
         if (a == one){
             continue;
         }

         powmod(a, s, p, m);

         for (j = ; j < d && !(m == one) && !(m == pd1); ++j){
             mulmod(m, m, p, m);
         }
         if (!(m == pd1) && j > ){
             return false;
         }
     }
     return true;
 }

 int main(void){
     bnum x;
     x.read();
     puts(MillerRabinTest(x, ) ? "Yes" : "No");
     return ;
 }

对于long long范围：

 #include <stdio.h>
 #include <iostream>
 #include <cstdlib>
 #define ll long long
 using namespace std;

 //利用二进制计算a*b%mod
 ll multi_mod(ll a, ll b, ll mod){
     ll ans = 0LL;
     a %= mod;
     while(b){
         if (b & 1LL) ans = (ans+a)%mod;
         b >>= 1LL;
         a = (a+a)%mod;
     }
     return ans;
 }

 //计算a^b%mod
 ll power_mod(ll a, ll b, ll mod){
     ll ans = 1LL;
     a %= mod;
     while(b){
         if(b & 1LL) ans = multi_mod(ans, a, mod);
         b >>= 1LL;
         a = multi_mod(a, a, mod);
     }
     return ans;
 }

 //Miller-Rabin测试,测试n是否为素数
 bool miller_rabin(ll n, int repeat){
     if (2LL == n || 3LL == n) return true;
     if (n%== || n%== || n%==) return false;

     //将n分解为2^s*d
     ll d = n - 1LL;
     int s = ;
     while(!(d & 1LL)){
         s++;
         d >>= 1LL;
     }
     // srand((unsigned)time(0));
     for(int i=; i<repeat; ++i){//重复repeat次
         ll a = rand() % (n - ) + ;//取一个随机数,[2,n-1)
         ll x = power_mod(a, d, n);
         ll y = 0LL;
         for(int j=; j<s; ++j){
             y = multi_mod(x, x, n);
             if (1LL == y && 1LL != x && n-1LL != x) return false;
             x = y;
         }
         if (1LL != y) return false;
     }
     return true;
 }

 int main(void){
     ll n;
     cin >> n;
     if(miller_rabin(n, )) cout << "Yes" << endl;
     else cout << "No" << endl;
     return ;
 }