建设一个大型网站大概费用,免费咨询医生在哪里咨询,创网络,扬中人才正题
题目链接:https://www.luogu.org/problemnew/show/P2842 题目大意
求 a#baaaa…a(a\#ba^{a^{a^{a^{…^a}}}}(a#baaaa…a(进行幂运算b次) 解题思路
根据费马小当p为质数时 ap≡a(modp)a^p\equiv a(mod\ p)ap≡a(mod p) 推导得ab≡ab%(p−1)(modp)a^b\equiv a^{b\%(p-1)}…正题
题目链接:https://www.luogu.org/problemnew/show/P2842 题目大意
求 a#baaaa…a(a\#ba^{a^{a^{a^{…^a}}}}(a#baaaa…a(进行幂运算b次) 解题思路
根据费马小当p为质数时 ap≡a(modp)a^p\equiv a(mod\ p)ap≡a(mod p) 推导得ab≡ab%(p−1)(modp)a^b\equiv a^{b\%(p-1)}(mod\ p)ab≡ab%(p−1)(mod p) 所以答案a#b≡aab−1%(p−1)(modp)a\# b\equiv a^{a^{b-1}\%(p-1)}(mod\ p)a#b≡aab−1%(p−1)(mod p) codecodecode
#includecstdio
#includealgorithm
#define ll long long
using namespace std;
const ll XJQ1e97;
ll a,b;
ll power(ll a,ll b,ll p)
{ll ans1;a%p;while(b){if(b1) ansans*a%p;aa*a%p;b1;}return ans;
}
int main()
{scanf(%lld%lld,a,b);ll kpower(a,b-1,XJQ-1);printf(%lld,power(a,k,XJQ));
}
