基准时间限制:1 秒 空间限制:131072 KB 分值: 40 难度:4级算法题
给出1个正整数N,检测N是否为质数。如果是,输出"Yes",否则输出"No"。
Input
输入一个数N(2 <= N <= 10^30)
Output
如果N为质数,输出"Yes",否则输出"No"。
Input示例
17
Output示例
Yes
大数的素数测试,套上模板
#include<iostream>
using namespace std;
//#include<stdlib>
#include<string>
#include<string.h>
#include<algorithm>
#define MAXL 4
#define M10 1000000000
#define Z10 9
const int zero[MAXL - 1] = {0};
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, 0, sizeof(data));
char buf[32];
scanf("%s", buf);
int len = (int)strlen(buf);
int i = 0, k;
while (len >= Z10)
{
for (k = len - Z10; k < len; ++k)
{
data[i] = data[i] * 10 + buf[k] - '0';
}
++i;
len -= Z10;
}
if (len > 0)
{
for (k = 0; k < len; ++k)
{
data[i] = data[i] * 10 + buf[k] - '0';
}
}
}
bool operator == (const bnum &x)
{
return memcmp(data, x.data, sizeof(data)) == 0;
}
bnum & operator = (const int x)
{
memset(data, 0, sizeof(data));
data[0] = x;
return *this;
}
bnum operator + (const bnum &x)
{
int i, carry = 0;
bnum ans;
for (i = 0; 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 = 0;
bnum ans;
for (i = 0; i < MAXL; ++i)
{
ans.data[i] = data[i] - x.data[i] - carry;
if (ans.data[i] < 0)
{
ans.data[i] += M10;
carry = 1;
}
else
{
carry = 0;
}
}
return ans;
}
// assume *this < x * 2
bnum operator % (const bnum &x)
{
int i;
for (i = MAXL - 1; i >= 0; --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 = 0, tmp;
for (i = MAXL - 1; i >= 0; --i)
{
tmp = data[i] & 1;
data[i] = (data[i] + carry) >> 1;
carry = tmp * M10;
}
return *this;
}
bool is_odd()
{
return (data[0] & 1) == 1;
}
bool is_zero()
{
for (int i = 0; i < MAXL; ++i)
{
if (data[i])
{
return false;
}
}
return true;
}
};
void mulmod(bnum &a0, bnum &b0, bnum &p, bnum &ans)
{
bnum tmp = a0, b = b0;
ans = 0;
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)
{
bnum tmp = a0, b = b0;
ans = 1;
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 = 0, j, d = 0;
for (i = 1; i < MAXL; ++i)
{
if (p.data[i])
{
break;
}
}
if (i == MAXL)
{
// small integer test
if (p.data[0] < 2)
{
return false;
}
if (p.data[0] == 2)
{
return true;
}
small = 1;
}
if (!p.is_odd())
{
return false; // even number
}
bnum a, s, m, one, pd1;
one = 1;
s = pd1 = p - one;
while (!s.is_odd())
{
s.div2();
++d;
}
for (i = 0; i < iter; ++i)
{
a = rand();
if (small)
{
a.data[0] = a.data[0] % (p.data[0] - 1) + 1;
}
else
{
a.data[1] = a.data[0] / M10;
a.data[0] %= M10;
}
if (a == one)
{
continue;
}
powmod(a, s, p, m);
for (j = 0; j < d && !(m == one) && !(m == pd1); ++j)
{
mulmod(m, m, p, m);
}
if (!(m == pd1) && j > 0)
{
return false;
}
}
return true;
}
int main()
{
bnum x;
x.read();
puts(MillerRabinTest(x, 5) ? "Yes" : "No");
return 0;
}