枚举每个置换，求在每个置换下着色不变的方法数，先求出每个循环的大小，再动态规划求得使用给定的颜色时对应的方法数。
dp[i][j][k]表示处理到当前圈时R,B，G使用量为i,j,k时的方法数，背包思想。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 68, INF = 0x3F3F3F3F;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
int r, g, b, m, p, n;
int dis[N][N];
bool vis[N];
LL dp[N][N][N]; LL Ext_gcd(LL a,LL b,LL &x,LL &y){//扩展欧几里得
if(b==0) { x=1, y=0; return a; }
LL ret= Ext_gcd(b,a%b,y,x);
y-= a/b*x;
return ret;
}
LL Inv(LL a,int m){ ///求逆元
LL d,x,y,t= (LL)m;
d= Ext_gcd(a,t,x,y);
if(d==1) return (x%t+t)%t;
return -1;
} LL solve(){
LL ans = 0;
for(int x = 0; x < m; x++){
memset(vis, 0, sizeof(vis));
vector<int> v;
for(int i = 1; i <= n; i++){
if(!vis[i]){
int cnt = 0;
int tp = i;
while(!vis[tp]){
cnt++;
vis[tp] = 1;
tp = dis[x][tp];
}
v.push_back(cnt);
}
}
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
for(int t = 0; t < v.size(); t++){
for(int i = r; i >= 0; i--){
for(int j = b; j >= 0; j--){
for(int k = g; k >= 0; k--){
if(i == 0 && j == 0 && k == 0){
continue;
}
dp[i][j][k] = 0;
if(i >= v[t]){
dp[i][j][k] = (dp[i][j][k] + dp[i - v[t]][j][k]) % p;
}
if(j >= v[t]){
dp[i][j][k] = (dp[i][j][k] + dp[i][j - v[t]][k]) % p;
}
if(k >= v[t]){
dp[i][j][k] = (dp[i][j][k] + dp[i][j][k - v[t]]) % p;
}
}
}
}
}
ans = (ans + dp[r][b][g]) % p;
}
ans = ans * Inv(m, p) % p;
return ans;
}
int main(){
while(~scanf("%d %d %d %d %d", &r, &b, &g, &m, &p)){
n = r + b + g;
bool f = 1;
for(int i = 0; i < m; i++){
int cnt = 0;
for(int j = 1; j <= n; j++){
scanf("%d", &dis[i][j]);
if(dis[i][j] == j){
cnt++;
}
}
if(cnt == n){
f = 0;
}
}
if(f){
for(int i = 1; i <= n; i++){
dis[m][i] = i;
}
m++;
}
printf("%lld\n", solve());
}
return 0;
}