HDU1261 大数

字串数
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3913 Accepted Submission(s): 950

Problem Description
一个A和两个B一共可以组成三种字符串:”ABB”,”BAB”,”BBA”.
给定若干字母和它们相应的个数,计算一共可以组成多少个不同的字符串.

Input
每组测试数据分两行,第一行为n(1<=n<=26),表示不同字母的个数,第二行为n个数A1,A2,…,An(1<=Ai<=12),表示每种字母的个数.测试数据以n=0为结束.

Output
对于每一组测试数据,输出一个m,表示一共有多少种字符串.

Sample Input
2
1 2
3
2 2 2
0

Sample Output
3
90

题意：
略
题解：
求总个数为多少，可以知道结果为(a1+a2+…..an)!/(a1!a2!*a3….an!),这里直接算肯定会超过数据上限，所以必须用高精度的大数。但是直接用高精度太慢又会超时。所以我们可以先预处理一下。设sum = a1+a2+…an。那么sum！=1*2*3…sum。我们先用一个数组a储存1到sum，然后对ai！= 1*2*3*…ai其中的每一个数求和数组a里每一个数的最大公约数，然后约去，将每个ai！约分到1，最终把得到的数组a里的每个数用大数乘起来就可以了。g++下耗时46ms

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <queue>#include <vector>#include <sstream>#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 a[400];int b[30];int gcd(int a,int b){    while(b^=a^=b^=a%=b);    return a;}int main(){    int n;    while(scanf("%d",&n)&&n) {        int sum = 0;        int i,j,k;        f(i,1,n){            scanf("%d",&b[i]);            sum+=b[i];        }        f(i,1,sum) a[i] = i;        f(i,1,n){            fi(j,b[i],2){                int t = j;                fi(k,sum,2){                    int tem = gcd(a[k],t);                    a[k]/=tem;                    t/=tem;                    if(t == 1)                        break;                }            }        }        INT ans = 1LL;        f(i,1,sum) ans*=a[i];        ans.print();        printf("\n");    }    return 0;}