【模板】欧拉定理
题目背景
模板题,无背景
题目描述
给你三个正整数,a,m,ba,m,ba,m,b,你需要求:
abmodma^b \mod mabmodm
输入输出格式
输入格式:一行三个整数,a,m,ba,m,ba,m,b
输出格式:一个整数表示答案
输入输出样例
说明
注意输入格式,a,m,ba,m,ba,m,b 依次代表的是底数、模数和次数
样例1解释:
24mod7=22^4 \mod 7 = 224mod7=2
输出2
数据范围:
对于全部数据:
1≤a≤1091≤a≤10^91≤a≤109
1≤b≤10200000001≤b≤10^{20000000}1≤b≤1020000000
1≤m≤1061≤m≤10^61≤m≤106
![](https://img2018.cnblogs.com/blog/1473307/201902/1473307-20190214093714305-44977519.png)
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<time.h> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) #define mclr(x,a) memset((x),a,sizeof(x)) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 98765431; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-5 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int a, m, phi = 1; int B, fg; ll qpow(ll x, ll y) { ll ans = 1; while (y) { if (y % 2)ans = ans * x%m; x = x * x%m; y >>= 1; } return ans; } int main() { // ios::sync_with_stdio(0); rdint(a); rdint(m); a %= m; int tmp = m; for (int i = 2; i <= sqrt(tmp); i++) { if (tmp%i)continue; phi *= (i - 1); tmp /= i; while (tmp%i == 0) { phi *= i; tmp /= i; } } if (tmp > 1)phi *= (tmp - 1); char ch; while ((ch = getchar()) < '0' || ch > '9'); while (B = B * 10ll + (ch^'0'), (ch = getchar()) >= '0'&&ch <= '9') { if (B >= phi)fg = 1, B %= phi; } if (B >= phi)fg = 1, B %= phi; if (fg)B += phi; printf("%lld\n", qpow(a * 1ll, B * 1ll)); return 0; }
EPFL - Fighting