n个1组成的整数能被2013整除

编写程序,求n至少为多大时,n个1组成的整数能被2013整除

#include<iostream>
using namespace std;
int main() {
    
    
	int n = 1, m = 1;
	do {
    
    
		n++;
		m = (m %2013) * 10 + 1;
	} while (m % 2013 != 0);
	cout << n << endl;
	return 0;
}

注:(m %2013) *10 + 1 要比 m *10 + 1 计算量小

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转载自blog.csdn.net/qq_45784099/article/details/114475224