$\huge \color{orange}{成魔之路->}$ $\huge \color{purple}{算法提高课题解}$

思路：

$点\ (n,m)\ 关于直线\ y=x+1\ 对称的点是\ (m-1,n+1)，答案=C_{n+m}^{n}-C_{n+m}^{n+1}$

$最后用高精度求解\ C_{n+m}^{n}-C_{n+m}^{n+1}$

完整代码

#include<bits/stdc++.h>

using namespace std;

const int N = 10010;

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

//线性筛质数
void init(int n) 
{
    for (int i = 2; i <= n; i++) 
    {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++) 
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

//返回 x! 中质因子 p 的个数
int get(int x, int p) 
{
    int res = 0;
    while (x) 
    {
        res += x / p;
        x /= p;
    }
    return res;
}

//高精度乘法
void mul(int r[], int &len, int p) 
{
    int t = 0;
    for (int i = 0; i < len; i++) 
    {
        t += r[i] * p;
        r[i] = t % 10;
        t /= 10;
    }
    while (t) 
    {
        r[len++] = t % 10;
        t /= 10;
    }
}

int C(int a, int b, int r[]) 
{
    int len = 1;
    r[0] = 1;
    for (int i = 0; i < cnt; i++) 
    {
        int p = primes[i];
        int s = get(a, p) - get(a - b, p) - get(b, p);
        while (s--) mul(r, len, p);
    }
    return len;
}

//高精度减法
void sub(int a[], int al, int b[], int bl) 
{
    for (int i = 0; i < al; i++) 
    {
        a[i] -= b[i];
        if (a[i] < 0) 
        {
            a[i] += 10;
            a[i + 1]--;
        }
    }
}

int main() 
{
    init(N - 1);

    cin >> n >> m;

    int al = C(n + m, n, a);
    int bl = C(n + m, n + 1, b);

    sub(a, al, b, bl);

    int k = al - 1;
    while (!a[k]) k--;
    while (k >= 0) cout << a[k--];

    return 0;
}