这道题我给大家三种解法

解法①：分配率

#include <iostream>
using namespace std;
int cnt(int n, int m, int k, int x)
{
	int ans = 1, a = 10;
	while(k)
    {
		if((k) & (1))
        {
            ans = ans * a % n;
        }	
		a = a * a % n;
		k >>= 1;
	}
	return (x % n + m % n * ans % n) % n;
}
int main()
{
	int n, m, k, x;
	cin >> n >> m >> k >> x;
    cout << cnt(n, m, k, x) << endl;
	return 0;
}

解法②：递增取余

#include <iostream>
using namespace std;
int main()
{
    long long n, m, k, x,y;
    cin >> n >> m >> k >> x;
    long long x1 = x, x2 = x;
    for (long long i = 0; i < 1e9; i++)
    {
        x += m;
        x %= n;
        if (x == x1)
        {
            x1 = i + 1;
            break;
        }
    }//x1圈后回到x位置
    int j = 1;
    for (int i = 0; i < k / 4; i++)
    {
        j *= 10000;
        j %= x1;
    }
    for (int i = 0; i < k % 4; i++)
    {
        j *= 10;
        j %= x1;
    }
    //剩余j圈
    for (int i = 0; i < j; i++)
    {
        x2 += m;
        x2 %= n;
    }
    cout << x2 << endl;
    return 0;
}

解法③：快速幂

#include <iostream>
using namespace std;
long long n, m, k, x;
int ksm(long long a, long long b, long long c)
{
	long long d = 1;
	a %= c;
	while (b > 0)
	{
		if (b % 2 == 1)
		    d = (d * a) % c;
		b /= 2;
		a = (a * a) % c;
	}
	return d;
}
int main()
{
	cin >> n >> m >> k >> x;
	k = ksm(10, k, n * m);
	k *= m, x += k;
	cout << x % n;
	return 0;
}