SoundHound Inc. Programming Contest 2018[C. Ordinary Beauty]

SoundHound Inc. Programming Contest 2018 -Masters Tournament-[C. Ordinary Beauty]

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打表找规律的。

1. \(n = 1\) 时， \(ans = m\)
2. \(n = 2\) 时， \(ans = 2*(m-1)*2^{m-2}\)

3. \(n = 3\) 时，
   1) \(d = 0, ~~ ans = 3*(m-1)*3^{m-2}\)
   2) \(d = 1, ~~ ans = 4*(m-1)*3^{m-2}\)
   3) \(d = 2, ~~ ans = 2*(m-1)*3^{m-2}\)

4. \(n = 4\) 时，
   1) \(d = 0, ~~ ans = 4*(m-1)*4^{m-2}\)
   2) \(d = 1, ~~ ans = 6*(m-1)*4^{m-2}\)
   3) \(d = 2, ~~ ans = 4*(m-1)*4^{m-2}\)
   4) \(d = 3, ~~ ans = 2*(m-1)*4^{m-2}\)
   发现：

   2 2
   3 4 2
   4 6 4 2
   5 8 6 4 2
   6 10 8 6 4 2

上图规律就显然了。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll n,m,d;
int main() {
    scanf("%lld%lld%lld",&n,&m,&d);
    double d0 = n, d1 = (n-1.0)*2.0;
    if(d==0){
        double ans = 1.0*d0*(m-1.0)/n/n;
        printf("%.8f\n",ans);
    }
    else if(d==1){
        double ans = 1.0*d1*(m-1.0)/n/n;
        printf("%.8f\n",ans);
    }
    else {
        double dd = d1 - 1.0*(d-1.0)*2.0;
        double ans = 1.0*dd*(m-1.0)/n/n;
        printf("%.8f\n",ans);
    }
    return 0;
}