CF460 A. Vasya and Socks

A. Vasya and Socks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every m-th day (at days with numbers m, 2m, 3m, ...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?

Input

The single line contains two integers n and m (1 ≤ n ≤ 100; 2 ≤ m ≤ 100), separated by a space.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input

2 2

output

3

input

9 3

output

13

Note

In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.

In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.

 

这道题真心不难，T T，没想全。。。没事，学到东西就好

我的想法

#include <iostream>
#include <cstdio>
using namespace std;
///自己的想法那么复杂，好吧，想法其实也可以，当时没好好想完全
///n > m没问题，而 n < m 没考虑 T-T
int main()
{
    int n, m;
    while(cin >> n >> m)
    {
        int ans = 0;

        do{
            if(n >= m)
            {
                ans += m;
                n = n - m + 1;
            }
            else
            {
                ans = n;
                n = 0;
            }
        }while(n >= m);
        ans += n;
        cout << ans << endl;
    }
    return 0;
}

参考，纯正模拟

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int n, m, i;
    cin >> n >> m;
    for(i = 1; n > 0; i++)
    {
        n--;
        n += (i%m == 0) ? 1 : 0;
    }
    //这种思想是完全模拟袜子的使用过程，应该就是这样的
    i--;
    cout << i << endl;
    return 0;
}