Game of Dice GYM101532E

http://codeforces.com/gym/101532/problem/E

meet in the middle

分成两份 然后后一份最终得到的A需要得到X时的B

A*B = X mod (1e9+7)
B = X*(inv[A]) mod (1e9+7)

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.10f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
const ll infl = 10000000000000000;
const int maxn=  100+10;
const int maxm = 6000+10;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}

int a[maxn][10];
ll qpow(ll k,ll x)
{
    ll res = 1;
    ll base = x;
    while(k)
    {
        if(k&1)res=res*base%mod;
        base = base*base %mod;
        k>>=1;
    }
    return res;
}
map<ll,int>ma;int n,x;
void dfs(int cur,int last,ll now)
{
    if(!last)
    {
        ma[now]++;
        return ;
    }
    for(int i=1;i<=6;++i)dfs(cur+1,last - 1, now*a[cur][i]%mod);

}
ll ans = 0;
void dfs2(int cur,ll now)
{
    if(cur==n+1)
    {
        ll inv = qpow(mod-2,now);
//        ll check = x%now==0?x/now:0;
//        ll cnt = ma[check];
//        ans +=cnt;
        inv = inv*x%mod ;
        ll cnt = ma[inv];
        ans += cnt;
        return ;
    }
    r1(i,6)dfs2(cur+1,now*a[cur][i]%mod);
}
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL


    int t;
    sd(t);
    while(t--)
    {
//        int n,x;
        ans  = 0;
        sdd(n,x);
        r1(i,n)r1(j,6)sd(a[i][j]);
        ma.clear();
        int mid = n/2;
        dfs(1,mid,1);
        dfs2(mid+1,1);
        lansn();
    }

    return 0;
}