1315. 网格 - AcWing题库

 

只要是触及上面这条红线的，就以第一次触及的点为起点沿红线反转，终点的位置与红线对称的位置可以看作触及红线的路线的终点。

y=x+1

横坐标容易得出时m - 1，（m + n） - (m - 1)得出纵坐标n + 1

答案就是

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
 
using namespace std;
 
typedef pair<int, int> PII;
typedef long long ll;
typedef long double ld;

const int N = 10010;

int n, m;
int primes[N], cnt;
bool st[N];

void init()
{
	for(int i = 2; i < N; i ++)
	{
		if(!st[i])primes[cnt ++] = i;
		for(int j = 0; primes[j] * i < N; j ++)
		{
			st[primes[j] * i] = true;
			if(i % primes[j] == 0)break;
		}
	}
}

int get(int n, int p)
{
	int c = 0;
	while(n)
	{
		c += n / p;
		n /= p;
	}
	return c;
}

vector<int> mul(vector<int> &A, int b)
{
	int t = 0;
	vector<int> C;
	for(int i = 0; i < A.size(); i ++)
	{
		t += A[i] * b;
		C.push_back(t % 10);
		t /= 10;
	}
	while(t)
	{
		C.push_back(t % 10);
		t /= 10;
	}
	return C;
}

vector<int> C(int a, int b)
{
	vector<int> A;
	A.push_back(1);
	for(int i = 0; primes[i] <= a; i ++)
	{
		int p = primes[i];
		int s = get(a, p) - get(b, p) - get(a - b, p);
		for(int j = 0; j < s; j ++)A = mul(A, p);
	}
	return A;
}

vector<int> sub(vector<int> &A, vector<int> &B)
{
	vector<int> c;
	int t = 0;
	for(int i = 0; i < A.size(); i ++)
	{
		t = A[i] - t;
		if(i < B.size())t -= B[i];
		c.push_back((t + 10) % 10);
		if(t < 0)t = 1;
		else t = 0;
	}
	while(c.size() > 1 && c.back() == 0)c.pop_back();
	return c; 
}

int main()
{
	IOS
	init();
	cin >> n >> m;
	vector<int> c1 = C(n + m, n), c2 = C(n + m, m - 1);
	c1 = sub(c1, c2);
	for(int i = c1.size() - 1; i >= 0; i --)
	{
		cout << c1[i];
	}
	
	return 0;
}