C++程序设计：Fast Mod Exponentiation快速模幂

Bob has encountered a difficult problem, and hope you design an algorithm to calculate pow(a,b) mod 1337, where a is a positive integer, b is a very large positive integer and will be given in the form of an array. For example, pow(2,3) mod 1337 is 8.

#include <iostream>
#include <vector>
using namespace std;

// pricinple: (a*b)%k = (a%k)(b%k)%k

//fast pow
long pow(long x, long n) {
	long curr=x,rs=1;
	while(n>0) {
		rs%=1337;
		curr%=1337;
		if(n%2) {
			rs*=curr;
		}
		curr*=curr;
		n/=2;
	}
	rs%=1337;
	return rs;
}

int dfs(int a, vector<int> b) {
	if(b.empty()) {
		return 1;
	}
	int last=b[b.size()-1];
	b.pop_back();
	return ((pow(a,last)%1337)*(pow(dfs(a,b),10)%1337)%1337);
}

int main() {
	int a;
	vector<int> b;
	string tmp;
	cin>>a;
	cin>>tmp;
	for(int i=0; i<tmp.length(); i++) {
		if(tmp[i]>='0' && tmp[i]<='9') {
			b.push_back(int(tmp[i]-'0'));
		}
	}
	cout<<dfs(a,b);
	return 0;
} 

标签：tmp,模幂,curr,Exponentiation,int,pow,1337,Fast,long
来源： https://blog.csdn.net/qq_41112170/article/details/120983638