解法一,全模拟

#include <iostream>
int main(){
    int m,k,day=0; std::cin>>m>>k;
    while(m>=k){
        if(day%k==0)m++;
        day+=1; m-=1;}
    std::cout<<day+m; return 0;}

解法二,每次过k天

#include <iostream>
int main(){
    int m,k,day=0; std::cin>>m>>k;
    while(m>=k){
        day+=k;
        m-=k-1;}
    std::cout<<day+m; return 0;}

解法三,计算

#include <iostream>
int main(){
    int m,k,day=0; std::cin>>m>>k;
    while(m>=k){
        day+=m/k*k;       //初步计算最多能跳x*k天
        m=m/k+m%k;}    //m等于m/k(满k所补贴的钱)+m%k(不满k,剩下的钱)
    std::cout<<day+m; return 0;}

计算量不大的话,推荐使用解法二

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int m,k,sum=0;
    cin >> m >> k ;
    if(k==1){
        cout << -1;
        return 0;
    }
    do{
        sum+=m;
        m/=k;
    }while(m);
    if(k==2){
        sum++;
    }
    cout << sum;
    return 0;
}

#include<iostream>
int main()
{
    int M,K;
    std::cin>>M>>K;
    int n=0;
    while(1)
    {
        M-=K;
        n+=K;
        M+=1;
        if(M<K)break;
    }
    n+=M;
    std::cout<<n;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    unsigned int m,k,sum=0;
    cin >> m;
    cin >> k;
    if(k==1){
        cout << -1;
        return 0;
    }
    do{
        sum+=m;
        m/=k;
    }while(m);
    if(k==2){
        sum++;
    }
    cout << sum;
    return 0;
}

带特殊情况的数学法计算，复杂度logn