逆元模板(根据费马小定理求逆元)
ll power(int a, long long b)//a为要求的逆元,b为mod-2的值
{
ll c = 1;
for (; b; b >>= 1)
{
if (b & 1)c = (long long)c*a%mod;
a = (long long)a*a%mod;
}
return c;
}
ll power(int a, long long b)//a为要求的逆元,b为mod-2的值
{
ll c = 1;
for (; b; b >>= 1)
{
if (b & 1)c = (long long)c*a%mod;
a = (long long)a*a%mod;
}
return c;
}
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