链接
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2222
题解
直接用burnside引理
置换群包含了旋转和翻转操作
分奇数偶数讨论一下,分别用组合数算一算不动点的数目就好了
代码
//polya
#include <bits/stdc++.h>
#define maxn 100
using namespace std;
typedef long long ll;
ll C[maxn][maxn], N;
ll read(ll x=0)
{
ll c, f=1;
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-48;
return f*x;
}
void init()
{
ll i, j;
for(i=0;i<maxn;i++)C[i][0]=1;
for(i=1;i<maxn;i++)for(j=1;j<=i;j++)C[i][j]=C[i-1][j-1]+C[i-1][j];
}
ll calc(ll size, ll a, ll b, ll c)
{
ll i;
if(a%size!=0 or b%size!=0 or c%size!=0)return 0;
a/=size, b/=size, c/=size;
return C[a+b+c][a]*C[b+c][b];
}
ll gcd(ll a, ll b){return !b?a:gcd(b,a%b);}
int main()
{
ll ans, a, b, c, n, i;
init();
N=read();
while(N--)
{
a=read(), b=read(), c=read();
n=a+b+c;
ans=0;
for(i=0;i<n;i++)ans+=calc(n/gcd(i,n),a,b,c);
if(n&1)
{
if(a)ans+=n*calc(2,a-1,b,c);
if(b)ans+=n*calc(2,a,b-1,c);
if(c)ans+=n*calc(2,a,b,c-1);
ans/=2*n;
}
else
{
ans+=n/2*calc(2,a,b,c);
if(a and b)ans+=n*calc(2,a-1,b-1,c);
if(a and c)ans+=n*calc(2,a-1,b,c-1);
if(b and c)ans+=n*calc(2,a,b-1,c-1);
if(a>1)ans+=n/2*calc(2,a-2,b,c);
if(b>1)ans+=n/2*calc(2,a,b-2,c);
if(c>1)ans+=n/2*calc(2,a,b,c-2);
ans/=2*n;
}
printf("%lld\n",ans);
}
return 0;
}