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
",ans);
    }
    else if(d==1){
        double ans = 1.0*d1*(m-1.0)/n/n;
        printf("%.8f
",ans);
    }
    else {
        double dd = d1 - 1.0*(d-1.0)*2.0;
        double ans = 1.0*dd*(m-1.0)/n/n;
        printf("%.8f
",ans);
    }
    return 0;
}