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HDU 4927

http://acm.hdu.edu.cn/showproblem.php?pid=4927

直接模拟会超时,要在纸上写写推公式

A[n]*C(0,n-1)  - A[n-1]*C(1,n-1) + A[n-2]*C(2,n-1)  - A[n-3]*C(3,n-1) ...... A[1]*C(n-1,n-1)

组合数可以用递推快速计算出来

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std; #define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigInt{ private:     int a[500];    //可以控制大数的位数     int len;       //大数长度public:     BigInt(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数    BigInt(const int);       //将一个int类型的变量转化为大数    BigInt(const char*);     //将一个字符串类型的变量转化为大数    BigInt(const BigInt &);  //拷贝构造函数    BigInt &operator=(const BigInt &);   //重载赋值运算符,大数之间进行赋值运算    friend istream& operator>>(istream&,  BigInt&);   //重载输入运算符    friend ostream& operator<<(ostream&,  BigInt&);   //重载输出运算符    BigInt operator+(const BigInt &) const;   //重载加法运算符,两个大数之间的相加运算     BigInt operator-(const BigInt &) const;   //重载减法运算符,两个大数之间的相减运算     BigInt operator*(const BigInt &) const;   //重载乘法运算符,两个大数之间的相乘运算     BigInt operator/(const int   &) const;    //重载除法运算符,大数对一个整数进行相除运算    BigInt operator^(const int  &) const;    //大数的n次方运算    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算        bool   operator>(const BigInt & T)const;   //大数和另一个大数的大小比较    bool   operator<(const BigInt & T) const;    bool   operator==(const BigInt & T) const;    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较    bool   operator<(const int &t) const;    bool   operator==(const int &t) const;     void print();       //输出大数}; bool BigInt::operator==(const BigInt & T) const {    return !(*this > T) && !(T > *this);}bool BigInt::operator==(const int &t) const {    BigInt T = BigInt(t);    return *this == T;}bool BigInt::operator<(const BigInt & T) const {    return T > *this;    }bool BigInt::operator<(const int &t) const {    return BigInt(t) > *this;}BigInt::BigInt(const int b)     //将一个int类型的变量转化为大数{     int c,d = b;    len = 0;    memset(a,0,sizeof(a));    while(d > MAXN)    {        c = d - (d / (MAXN + 1)) * (MAXN + 1);         d = d / (MAXN + 1);        a[len++] = c;    }    a[len++] = d;}BigInt::BigInt(const char*s)     //将一个字符串类型的变量转化为大数{    int t,k,index,l,i;    memset(a,0,sizeof(a));    l=strlen(s);       len=l/DLEN;    if(l%DLEN)        len++;    index=0;    for(i=l-1;i>=0;i-=DLEN)    {        t=0;        k=i-DLEN+1;        if(k<0)            k=0;        for(int j=k;j<=i;j++)            t=t*10+s[j]-0;        a[index++]=t;    }}BigInt::BigInt(const BigInt & T) : len(T.len)  //拷贝构造函数{     int i;     memset(a,0,sizeof(a));     for(i = 0 ; i < len ; i++)        a[i] = T.a[i]; } BigInt & BigInt::operator=(const BigInt & n)   //重载赋值运算符,大数之间进行赋值运算{    int i;    len = n.len;    memset(a,0,sizeof(a));     for(i = 0 ; i < len ; i++)         a[i] = n.a[i];     return *this; }istream& operator>>(istream & in,  BigInt & b)   //重载输入运算符{    char ch[MAXSIZE*4];    int i = -1;    in>>ch;    int l=strlen(ch);    int count=0,sum=0;    for(i=l-1;i>=0;)    {        sum = 0;        int t=1;        for(int j=0;j<4&&i>=0;j++,i--,t*=10)        {            sum+=(ch[i]-0)*t;        }        b.a[count]=sum;        count++;    }    b.len =count++;    return in;}ostream& operator<<(ostream& out,  BigInt& b)   //重载输出运算符{    int i;      cout << b.a[b.len - 1];     for(i = b.len - 2 ; i >= 0 ; i--)    {         cout.width(DLEN);         cout.fill(0);         cout << b.a[i];     }     return out;}BigInt BigInt::operator+(const BigInt & T) const   //两个大数之间的相加运算{    BigInt t(*this);    int i,big;      //位数       big = T.len > len ? T.len : len;     for(i = 0 ; i < big ; i++)     {         t.a[i] +=T.a[i];         if(t.a[i] > MAXN)         {             t.a[i + 1]++;             t.a[i] -=MAXN+1;         }     }     if(t.a[big] != 0)        t.len = big + 1;     else        t.len = big;       return t;}BigInt BigInt::operator-(const BigInt & T) const   //两个大数之间的相减运算 {      int i,j,big;    bool flag;    BigInt t1,t2;    if(*this>T)    {        t1=*this;        t2=T;        flag=0;    }    else    {        t1=T;        t2=*this;        flag=1;    }    big=t1.len;    for(i = 0 ; i < big ; i++)    {        if(t1.a[i] < t2.a[i])        {             j = i + 1;             while(t1.a[j] == 0)                j++;             t1.a[j--]--;             while(j > i)                t1.a[j--] += MAXN;            t1.a[i] += MAXN + 1 - t2.a[i];         }         else            t1.a[i] -= t2.a[i];    }    t1.len = big;    while(t1.a[t1.len - 1] == 0 && t1.len > 1)    {        t1.len--;        big--;    }    if(flag)        t1.a[big-1]=0-t1.a[big-1];    return t1; } BigInt BigInt::operator*(const BigInt & T) const   //两个大数之间的相乘运算 {     BigInt ret;     int i,j,up;     int temp,temp1;       for(i = 0 ; i < len ; i++)    {         up = 0;         for(j = 0 ; j < T.len ; j++)        {             temp = a[i] * T.a[j] + ret.a[i + j] + up;             if(temp > MAXN)            {                 temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);                 up = temp / (MAXN + 1);                 ret.a[i + j] = temp1;             }             else            {                 up = 0;                 ret.a[i + j] = temp;             }         }         if(up != 0)             ret.a[i + j] = up;     }     ret.len = i + j;     while(ret.a[ret.len - 1] == 0 && ret.len > 1)        ret.len--;     return ret; } BigInt BigInt::operator/(const int & b) const   //大数对一个整数进行相除运算{     BigInt ret;     int i,down = 0;       for(i = len - 1 ; i >= 0 ; i--)    {         ret.a[i] = (a[i] + down * (MAXN + 1)) / b;        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;    }     ret.len = len;     while(ret.a[ret.len - 1] == 0 && ret.len > 1)        ret.len--;     return ret; }int BigInt::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算    {    int i,d=0;    for (i = len-1; i>=0; i--)    {        d = ((d * (MAXN+1))% b + a[i])% b;      }    return d;}BigInt BigInt::operator^(const int & n) const    //大数的n次方运算{    BigInt t,ret(1);    int i;    if(n<0)        exit(-1);    if(n==0)        return 1;    if(n==1)        return *this;    int m=n;    while(m>1)    {        t=*this;        for( i=1;i<<1<=m;i<<=1)        {            t=t*t;        }        m-=i;        ret=ret*t;        if(m==1)            ret=ret*(*this);    }    return ret;}bool BigInt::operator>(const BigInt & T) const   //大数和另一个大数的大小比较{     int ln;     if(len > T.len)        return true;     else if(len == T.len)    {         ln = len - 1;         while(a[ln] == T.a[ln] && ln >= 0)            ln--;         if(ln >= 0 && a[ln] > T.a[ln])            return true;         else            return false;     }     else        return false; }bool BigInt::operator >(const int & t) const    //大数和一个int类型的变量的大小比较{    BigInt b(t);    return *this>b;}void BigInt::print()    //输出大数{     int i;       cout << a[len - 1];     for(i = len - 2 ; i >= 0 ; i--)    {         cout.width(DLEN);         cout.fill(0);         cout << a[i];     }     cout << endl;}BigInt a[3005] ;BigInt C[3005] ;int main(){    int T ;    cin >> T ;    while(T--)    {        int n ;        cin >> n ;        for(int i=0 ;i<n ;i++)            cin >> a[i] ;        C[0]=1 ;        for(int i=1 ;i<n ;i++)            C[i]=C[i-1]*(n-i)/i ;        BigInt ans1,ans2 ;        ans1=ans2=0 ;        for(int i=0 ;i<n ;i++)        {            if((n-i)&1)ans1=ans1+a[i]*C[i] ;            else ans2=ans2+a[i]*C[i] ;         }         (ans1-ans2).print() ;    }    return 0; }
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HDU 4927