P2613有理数取余

P2613 一次同余式+费马小定理

处理输入(字符串)

一次同余式：\(ax≡b(mod p)\)

(1) \(gcd(a,p)=1\)，\(b\)任意，唯一解

(2) \(gcd(a,p)=d>1\)，\(d\nmid b\)，无解

(3) \(gcd(a,p)=d>1，d\mid b\)，d个解

这题19260817是质数，可以快速由\(gcd(a,p)=?\)，处理掉(2)和(3)

只剩唯一解，\(x ≡ b · a^{-1} (mod p)\)，\(a^{-1}\)用快速幂计算(当\(gcd(a,p)=1\)时，有 \(a^{p-2}≡ a^{-1} (mod p)\))

注意：题目是 \(bx≡a\)

---

上述结论中，\(b=1\)时，

\(ax≡1(mod p)\)

(1) \(a,p\)互质，唯一解

(2) \(a,p\)不互质，无解

---

题意：

求\(ans=a \times b^{-1}\)

思路：

这题的输入很大，需要用字符串存入，逐位读取，边读边取模

读完之后，得到的整数b(<=mod)，可以用费马小定理求逆元

---

const int mod=19260817;
string a,b;
int ia,ib;

int qpow(int a,int b)
{
    int ans=1;
    while(b){
        if(b&1) ans=1ll*ans*a%mod;
        a=1ll*a*a%mod;
        b>>=1;
    }
    return ans;
}

int main()
{
    cin>>a>>b;
    for(char& ch:a){
        ia=(ia*10+ch-'0')%mod;
    }
    for(char& ch:b){
        ib=(ib*10+ch-'0')%mod;
    }
    if(ib%mod==0) cout<<"Angry!";
    else{
        //gcd(ib,mod)=1,费马小定理求逆元
        int inv_b=qpow(ib,mod-2);
        cout<<1ll*ia*inv_b%mod;
    }
    return 0;
}

测试点

输入：

1
1926081700000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

输出：

Angry!