【CF963C】Cutting Rectangle（数论，构造，map）

题意：

 思路：考虑构造最小的单位矩形然后平铺

单位矩形中每种矩形的数量可以根据比例算出来，为c[i]/d，其中d是所有c[i]的gcd，如果能构造成功答案即为d的因子个数

考虑如果要将两种矩形放在同一行那他们的w一定相等，且对于每一行h全部出现过并且比例相当

具体实现的时候用map套vector，发现map有好多写法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef pair<ll,ll>P;
#define N  2000000+10
#define M  200000+10
#define INF 1e9
#define fi first
#define se second
#define MP make_pair
#define pb push_back
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1
#define fors(i) for(auto i:e[x]) if(i!=p)

const int MOD=998244353,inv2=(MOD+1)/2;
      //int p=1e4+7;
      //double eps=1e-6;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

struct node
{
    ll w,h,c;
}a[N];

map<ll,vector<Pll>> mp;
vector<Pll> c;

ll readll()
{
   ll v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

ll gcd(ll x,ll y)
{
    if(y==0) return x;
    return gcd(y,x%y);
}

int isok()
{
    for(auto &element:mp)
    {
        if(element.se.size()!=c.size()) return 0;
        int n=c.size();
        ll d=0;
        rep(i,0,n-1) d=gcd(d,element.se[i].se);
        rep(i,0,n-1)
        {
            element.se[i].se/=d;
            if(element.se[i]!=c[i]) return 0;
        }
    }
    return 1;
}

int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int n=read();
    ll K=0;
    rep(i,1,n)
    {
        a[i].w=readll(),a[i].h=readll(),a[i].c=readll();
        K=gcd(K,a[i].c);
        mp[a[i].w].pb(Pll(a[i].h,a[i].c));
    }
    for(auto &element:mp)
    {
        for(auto &x:element.se) x.se/=K;
        sort(element.se.begin(),element.se.end());
    }

    c=mp.begin()->se;
    ll d=0;
    for(auto &x:c) d=gcd(d,x.se);
    for(auto &x:c) x.se/=d;
    if(!isok())
    {
        printf("0
");
        return 0;
    }
    int ans=0;
    int t=sqrt(K);
    rep(i,1,t)
     if(K%i==0)
     {
         ans++;
         if(1ll*i*i!=K) ans++;
     }
    printf("%d
",ans);
    return 0;
}