[Day2] cinema ticket

cinema ticket

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Their thinking

The easy recursive formula is $f_{i,j}=f_{i-1,j}+f_{i,j-1}$ .
When known $i==j$ ,$f_{i,j}$ For the number of catterland $i$ term .
You can push $ans=\frac{(n+1-m)(n+m)!}{m!(n+1)!}$ .
High precision is needed .

code

#include<iostream>
#include<cstdio>
using namespace std;

int n,m;

struct num{
    
	int s[10010];
	num operator *(int x)
	{
    
		num t={
    0};
		int lst=0;
		for(int i=10000;i;i--)
		{
    
			t.s[i]=s[i]*x+lst;
			lst=t.s[i]/10;
			t.s[i]%=10;
		}
		return t;
	}
	num operator /(int x)
	{
    
		num t={
    0};
		int lst=0;
		for(int i=1;i<=10000;i++)
			lst=lst*10+s[i],t.s[i]=lst/x,lst%=x;
		return t;
	}
	num operator -(num x)
	{
    
		num t={
    0};
		for(int i=10000;i;i--)
		{
    
			if(s[i]>=x.s[i])
				t.s[i]=s[i]-x.s[i];
			else
				t.s[i]=s[i]+10-x.s[i],s[i-1]--;
		}
		return t;
	}
}a,b;

int main()
{
    
	cin>>n>>m;
	int i,j;
	b.s[10000]=1;
	for(i=1,j=n+m;i<m;i++,j--)
		b=b*j,b=b/i;
	a=b*j,a=a/i;
	num ans=a-b;
	i=0;
	while(!ans.s[i]) i++;
	while(i<=10000) cout<<ans.s[i++];
}