Eva

Description

某城市的街道呈网格状，左下角坐标为A(0, 0)，右上角坐标为B(n, m)，其中n >= m。现在从A(0, 0)点出发，只能沿着街道向正右方或者正上方行走，且不能经过图示中直线左上方的点，即任何途径的点(x, y)都要满足x >= y，请问在这些前提下，到达B(n, m)有多少种走法。

Input

输入文件中仅有一行，包含两个整数n和m，表示城市街区的规模。

Output

输出文件中仅有一个整数和一个换行/回车符，表示不同的方案总数。

Sample Input

6 6

Sample Output

132

HINT 

100%的数据中，1 <= m <= n <= 5 000

Source

Solution

这题n==m时是组合数

然后正解是类似于catalan数的证明方法C(n+m,n)-C(n+m,n+1)

发一发高精度的板子QAQ

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int mt=10000+5;
struct bign
{
    int len,s[mt];
    bign ()
    {
        memset(s, 0, sizeof(s)),len = 1;
    }
    bign (int num) { *this = num; }
    bign (const char *num) { *this = num; }
    bign operator = (const int num)
    {
        char s[mt];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num)
    {
        for(int i = 0; num[i] == '0'; num++) ;
        len = strlen(num);
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
    bign operator + (const bign &b) const
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    void clean()
    {
        while(len > 1 && !s[len-1]) len--;
    }
    bign operator * (const bign &b)
    {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
        {
            for(int j = 0; j < b.len; j++)
            {
                c.s[i+j] += s[i] * b.s[j];
            }
        }
        for(int i = 0; i < c.len; i++)
        {
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator - (const bign &b)
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator / (const bign &b)
    {
        bign c, f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10;
            f.s[0] = s[i];
            while(f >= b)
            {
                f =f-b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bool operator < (const bign &b)
    {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        }
        return false;
    }
    bool operator > (const bign &b)
    {
        if(len != b.len) return len > b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        }
        return false;
    }
    bool operator == (const bign &b)
    {
        return !(*this > b) && !(*this < b);
    }
    bool operator != (const bign &b)
    {
        return !(*this == b);
    }
    bool operator <= (const bign &b)
    {
        return *this < b || *this == b;
    }
    bool operator >= (const bign &b)
    {
        return *this > b || *this == b;
    }
    string str() const
    {
        string res = "";
        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
        return res;
    }
};
istream& operator >> (istream &in, bign &x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, const bign &x)
{
    out << x.str();
    return out;
}
int main()
{
    bign a,b,d=1,e;
    cin>>a>>b;
    bign c=a+b,y=1;
    for(bign i=a+y;i<=c;i=i+y) d=d*i;
    for(bign i=y;i<=b;i=i+y) d=d/i;
    e=d*b/(a+y);
    bign ans=d-e;
    ans.clean();
    cout<<ans<<endl;
    return 0;
}

/**************************************************************

    Problem: 3907

    User: ictsing

    Language: Python

    Result: Accepted

    Time:34 ms

    Memory:8200 kb

****************************************************************/

 

fac={};

def C(n,m):

    return fac[n]/fac[m]/fac[n-m];

fac[0]=1;

for i in range(1,10000):

    fac[i]=fac[i-1]*i;

f=raw_input().split(" ");

n=int(f[0]);m=int(f[1]);

print(C(n+m,n)-C(n+m,n+1));