POJ 1942 组合数学

Paths on a Grid
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 23661 Accepted: 5826

Description

Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he’s explaining that (a+b)2=a2+2ab+b2). So you decide to waste your time with drawing modern art instead.

Fortunately you have a piece of squared paper and you choose a rectangle of size n*m on the paper. Let’s call this rectangle together with the lines it contains a grid. Starting at the lower left corner of the grid, you move your pencil to the upper right corner, taking care that it stays on the lines and moves only to the right or up. The result is shown on the left:

Really a masterpiece, isn’t it? Repeating the procedure one more time, you arrive with the picture shown on the right. Now you wonder: how many different works of art can you produce?
Input

The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m, denoting the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension. Input is terminated by n=m=0.
Output

For each test case output on a line the number of different art works that can be generated using the procedure described above. That is, how many paths are there on a grid where each step of the path consists of moving one unit to the right or one unit up? You may safely assume that this number fits into a 32-bit unsigned integer.
Sample Input

5 4
1 1
0 0

Sample Output

126
2

题意：
有m*n个方格，问从左下角走到右上角一共有多少种走法。
题解
C（m+n，m）直接算，用上大数一次过。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <queue>#include <sstream>#include <vector>#define f(i,a,b) for(i = a;i<=b;i++)#define fi(i,a,b) for(i = a;i>=b;i--)using namespace std;#define LEN 500#define MOD 10000//定义了+ - * / % = == != >> << < > <= >= += -= *= /= %= print length pow等操作struct INT{    int num[LEN], len;    bool sign;    inline INT(long long x = 0)    {        *this = x;    }    inline INT(const string &str)    {        *this = str;    }    inline INT(const int a[], int b, bool c)    {        memcpy(num, a, sizeof num);        len = b; sign = c;    }    inline INT &operator =(const string &str)    {        int start = 0;        len = 0; sign = false;        memset(num, 0, sizeof num);        if (str[0] == '-') sign = true, start = 1;        while (str[start] == '0') start++;        for (int i = str.length() - 1; i >= start; i -= 4, len++)            for (int j = max(start, i - 3); j <= i; j++)                num[len] = (num[len] << 3) + (num[len] << 1) + str[j] - '0';        if (!len) sign = false;        if (len) len--;        return *this;    }    inline INT &operator =(long long x)    {        len = 0; sign = false;        memset(num, 0, sizeof num);        if (x < 0) sign = true, x = -x;        while (x)            num[len++] = x % MOD,                         x /= MOD;        if (len) len--;        return *this;    }    inline int length() const    {        int re = len << 2, t = num[len];        while (t) t /= 10, re++;        return re;    }    inline void print()    {        if (sign) putchar('-');        printf("%d", num[len]);        for (int i = len - 1; i >= 0; i--)            printf("%04d", num[i]);    }    inline friend void print_to_string(const INT &x, string &y)    {        stringstream stream;        stream << x;        stream >> y;    }    inline friend INT pow(const INT &x, int y)    {        INT re = 1, _x = x;        while (y)        {            if (y & 1)                re *= _x;            y >>= 1;            _x *= _x;        }        return re;    }    inline friend INT pow(const INT &x, const INT &y)    {        INT re = 1, _x = x, _y = y;        while (_y != 0)        {            if (_y.num[0] & 1)                re *= _x;            _y = shr(_y);            _x *= _x;        }        return re;    }    inline friend istream &operator >>(istream &in, INT &x)    {        string str;        in >> str;        x = str;        return in;    }    inline friend ostream &operator <<(ostream &out, const INT &x)    {        if (x.sign) out << '-';        out << x.num[x.len];        for (int i = x.len - 1; i >= 0; i--)            out.fill('0'), out.width(4), out << x.num[i];        return out;    }    inline INT operator -() const    {        return INT(num, len, !sign);    }    inline friend INT abs(const INT &x)    {        return INT(x.num, x.len, false);    }    inline friend bool operator <(const INT &x, const INT &y)    {        if (x.sign ^ y.sign) return x.sign;        int lx = x.length(), ly = y.length();        if (lx == ly)        {            for (int i = x.len; i >= 0; i--)                if (x.num[i] != y.num[i])                    return (x.num[i] < y.num[i])^x.sign;            return false;        }        return (lx < ly)^x.sign;    }    inline friend bool operator >(const INT &x, const INT &y) { return y < x; }    inline friend bool operator <=(const INT &x, const INT &y) { return !(y < x); }    inline friend bool operator >=(const INT &x, const INT &y) { return !(x < y); }    inline friend bool operator ==(const INT &x, const INT &y) { return !(x < y || y < x); }    inline friend bool operator !=(const INT &x, const INT &y) { return !(x == y); }    inline friend INT operator +(const INT &x, const INT &y)    {        if (x.sign ^ y.sign)            return x - (-y);        INT re;        re.sign = x.sign;        re.len = max(x.len, y.len);        for (int i = 0; i <= re.len; i++)        {            re.num[i] += x.num[i] + y.num[i];            re.num[i + 1] = re.num[i] / MOD;            re.num[i] %= MOD;        }        if (re.num[re.len + 1]) re.len++;        return re;    }    inline friend INT operator -(const INT &x, const INT &y)    {        if (x.sign ^ y.sign)            return x + (-y);        INT re, _x = x, _y = y;        re.sign = _x < _y;        if (re.sign ^ _x.sign)            swap(_x, _y);        for (int i = 0; i <= _x.len; i++)        {            re.num[i] += _x.num[i] - _y.num[i];            if (re.num[i] < 0)                re.num[i] += MOD,                             re.num[i + 1]--;        }        re.len = _x.len;        while (!re.num[re.len] && re.len >= 0) re.len--;        return re;    }    inline friend INT operator *(const INT &x, const INT &y)    {        INT re, _x = x, _y = y;        while (_y != 0)        {            if (_y.num[0] & 1)                re += _x;            _y = shr(_y);            _x += _x;        }        if (y.sign) re.sign ^= 1;        return re;    }    inline friend INT operator /(const INT &x, const INT &y)    {        if ((!y.len && !y.num[0]) || (!x.len && !x.num[0]) || abs(x) < abs(y)) { return INT(); }        INT re, left, _y = abs(y);        re.sign = x.sign ^ y.sign;        re.len = x.len - y.len + 1;        left.len = -1;        for (int i = x.len; i >= 0; i--)        {            memmove(left.num + 1, left.num, sizeof(left.num) - sizeof(int));            left.len++;            left.num[0] = x.num[i];            int l = 0, r = MOD - 1, mid;            if (left < y) r = 1;            while (l < r)            {                mid = (l + r) >> 1;                INT t = mid;                if (t * _y <= left)                    l = mid + 1;                else r = mid;            }            re.num[i] = r - 1;            INT t = r - 1;            left = left - (t * _y);        }        while (re.num[re.len] == 0 && re.len) re.len--;        return re;    }    inline friend INT operator %(const INT &x, const INT &y)    {        if ((!y.len && !y.num[0]) || (!x.len && !x.num[0])) { return INT(); }        INT left, _y = abs(y);        left.sign = (x.sign && !y.sign);        left.len = -1;        for (int i = x.len; i >= 0; i--)        {            memmove(left.num + 1, left.num, sizeof(left.num) - sizeof(int));            left.len++;            left.num[0] = x.num[i];            int l = 0, r = MOD - 1, mid;            while (l < r)            {                mid = (l + r) >> 1;                INT t = mid;                if (t * _y <= left)                    l = mid + 1;                else r = mid;            }            INT t = r - 1;            left = left - (t * _y);        }        return left;    }    inline friend INT shr(const INT &x)    {        INT re;        re.len = x.len;        for (int i = re.len; i >= 0; i--)        {            if (x.num[i] & 1 && i - 1 >= 0)                re.num[i - 1] += MOD >> 1;            re.num[i] += x.num[i] >> 1;        }        if (re.len && !re.num[re.len]) re.len--;        return re;    }    INT &operator +=(const INT &x) { return *this = *this + x; }    INT &operator -=(const INT &x) { return *this = *this - x; }    INT &operator *=(const INT &x) { return *this = *this * x; }    INT &operator /=(const INT &x) { return *this = *this / x; }    INT &operator %=(const INT &x) { return *this = *this % x; }};int main(){    long long n,m;    while(~scanf("%lld%lld",&n,&m)){        if(0==n&&0==m)            break;        INT k = n;        if(m<n) k = m;        INT a = 1LL;        long long i = m+n,j;        f(j,1,k) a*=i--;        INT b = 1LL;        f(j,1,k) b*=j;        cout << a/b << endl;    }    return 0;}