FZU 1759-Super A^B mod C(快速冪+大整數模數+歐拉函數)

標籤：blog   http   io   ar   os   sp   for   on   2014   

題目連結：點擊開啟連結

題意：計算 a^b %c 但其中b很大，可能會達到10^1000000, 故有降冪公式 a^b %c= a^(b%phi(c)+phi(c)) %c  （b>=phi(c)） 

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cctype>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>#define maxn 10100#define _ll __int64#define ll long long#define INF 0x3f3f3f3f#define Mod 1000000007#define pp pair<int,int>#define ull unsigned long longusing namespace std;int a,c;char b[1000010];ll phi(ll n){ll m=(ll)sqrt(n+0.5),ans=n;for(ll i=2;i<=m;i++){if(n%i==0){ans=ans/i*(i-1);while(n%i==0)n/=i;}}if(n>1)ans=ans/n*(n-1);return ans;}ll pow_mod(ll a,ll n,ll p){if(n==0)return 1;ll ans=pow_mod(a,n/2,p);ans=ans*ans%p;if(n&1)ans=ans*a%p;return ans;}void solve(){int len=strlen(b);ll tem=phi(c),sb;if(len<=10){sscanf(b,"%I64d",&sb);if(sb>=tem)printf("%I64d\n",pow_mod(a,sb%tem+tem,c));elseprintf("%I64d\n",pow_mod(a,sb,c));return ;}ll ans=0;for(int i=0;i<len;i++)ans=(ans*10+(b[i]-'0'))%tem;printf("%I64d\n",pow_mod(a,ans+tem,c));}int main(){while(~scanf("%I64d %s %I64d",&a,b,&c))solve();    return 0;}

FZU 1759-Super A^B mod C(快速冪+大整數模數+歐拉函數)