HUST 1372 marshmallow

Posted 2020-06-16

很简单的博弈题.....算几组能得到规律了。

某个状态先手要赢 等价于 之前有一种状态是后手赢，先手可以保证让现在这个状态到达那个状态

#include<cstdio>
#include<cstring>
#include<ctime>
#include<algorithm>
using namespace std;

const int maxn = 10000 + 10;
int a[maxn], b[maxn];
int n, m;

int gcd(int a, int b)
{
    int t;
    while (b)
    {
        t = a%b;
        a = b;
        b = t;
    }
    return a;
}

void init()
{
    for (int i = 1; i <= 10000; i++) a[i] = i, b[i] = i;
    for (int i = 3; i <= 10000; i = i + 3)
    {
        swap(b[i - 1], b[i - 2]);
    }
    b[10000] = 10001;

}

int main()
{
    init();
    while (~scanf("%d%d", &n, &m)){
        int tot = 0;
        for (int i = 1; i <= 10000; i++)
            if (a[i] <= n&&b[i] <= m) tot++;

        int fz, fm;
        fz = tot;
        fm = m*n;
        if (tot == 0) fz = 0, fm = 1;
        else
        {
            int a=fz / gcd(fz, fm),b = fm / gcd(fz, fm);
            fz = a; fm = b;
        }
        printf("%d/%d\n", fz, fm);
    }
    return 0;
}