标签：++ euler include 循环 printf show ima sqrt div
题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2837
题目大意：
已知f(0) = 1,0^0 =1,【注意，0的其他任意次方为0，虽然题没有直接给出~】，还已知f(n) = (n%10)^f(n/10),让你求f(n)%m. (2 ≤ n , m ≤ 10^9）
解题思路：
通过这个就可以递归求解。
f(n) = f(n%10)(f(n/10)%Phi(m)+Phi(m))%m
递归返回值就是f(n)%m+m
注意等于0的细节部分
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 using namespace std; 6 typedef long long ll; 7 int euler_phi(int n)//求单个 8 { 9 int m = (int)sqrt(n + 0.5); 10 int ans = n; 11 for(int i = 2; i <= m; i++)if(n % i == 0) 12 { 13 ans = ans / i * (i - 1); 14 while(n % i == 0)n /= i; 15 } 16 if(n > 1)ans = ans / n * (n - 1); 17 return ans; 18 } 19 ll mul(ll a, ll b, ll m) 20 //求a*b%m 21 { 22 ll ans = 0; 23 a %= m; 24 while(b) 25 { 26 if(b & 1)ans = (ans + a) % m; 27 b /= 2; 28 a = (a + a) % m; 29 } 30 return ans; 31 } 32 ll pow(ll a, ll b, ll m) 33 { 34 ll ans = 1; 35 a %= m; 36 while(b) 37 { 38 if(b & 1)ans = mul(a, ans, m); 39 b /= 2; 40 a = mul(a, a, m); 41 } 42 ans %= m; 43 return ans; 44 } 45 ll f(ll n, ll m) 46 { 47 if(n < 10)return n; 48 ll p = euler_phi(m); 49 ll t = f(n / 10, p); 50 ll ans = pow(n % 10, t, m); 51 if(n % 10 == 0 && t != 0)return 0;//直接等于0 52 if(ans == 0)ans = m;//返回值为f(n)%m+m 53 return ans; 54 } 55 int main() 56 { 57 int T; 58 scanf("%d", &T); 59 while(T--) 60 { 61 ll n, m; 62 scanf("%lld%lld", &n, &m); 63 printf("%lld\n", f(n, m) % m); 64 } 65 return 0; 66 }
标签：++ euler include 循环 printf show ima sqrt div
原文地址：https://www.cnblogs.com/fzl194/p/9074719.html
