P2613 【模板】有理数取余(逆元求除法的模+费马小定理/扩展欧几里得)

传送门

题目描述:

给出一个有理数 c=a/b​，求 c mod 19260817 的值。

思路:求b关于mod的逆元k,得(k*b)%m=1,(a/b)mod m=

((a/b)%m)*(k*b%m)=a*k%mod,a,b的数值很大,需要在读数

的时候用快读模板对m求余先,逆元这题除了

扩展欧几里得:

void extend_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1, y = 0;
        return;
    }
    extend_gcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a / b) * y;
}
ll mod_rev(ll a, ll m) {
    ll x, y;
    extend_gcd(a, m, x, y);
    return (m + x % m) % m;
}

 

这题因为m是素数,与a互素,可以用费马小定理,

a^（m-1）≡1（mod m),则a'=pow(a,m-2);

ll qpow(ll a, ll n) {
    ll ans = 1;
    while (n) {
        if (n & 1) {
            ans = (ans * a) % mod;
        }
        a = (a * a) % mod;
        n >>= 1;
    }
    return ans;
}
ll mod_rev2(ll a, ll m) {
    return qpow(a, m - 2);
}

AC代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 10000005;
const int inf = 0x3f3f3f3f;
const int mod = 19260817;
void read(ll& sum) {
    sum = 0; char p = getchar();
    for (; !isdigit(p); p = getchar());
    for (; isdigit(p); p = getchar()) {
        sum = sum * 10 % mod + p - '0';
    }
    sum% mod;
}
void extend_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1, y = 0;
        return;
    }
    extend_gcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a / b) * y;
}
ll mod_rev(ll a, ll m) {
    ll x, y;
    extend_gcd(a, m, x, y);
    return (m + x % m) % m;
}
ll qpow(ll a, ll n) {
    ll ans = 1;
    while (n) {
        if (n & 1) {
            ans = (ans * a) % mod;
        }
        a = (a * a) % mod;
        n >>= 1;
    }
    return ans;
}
ll mod_rev2(ll a, ll m) {
    return qpow(a, m - 2);
}
ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
}
int main() {
    //freopen("test.txt", "r", stdin);
    ll a, b;
    read(a);//大数需要对m求余
    read(b);
    if (gcd(b, 19260817) != 1) {
        printf("Angry!\n"); return 0;
    }
    printf("%lld\n", a * mod_rev2(b, 19260817) % 19260817);
    return 0;
}