hdu1134 Game of Connections(Catalan数， 顺便附上大数类模板)

Catalan数公式：

C[0] = 1

C[n] = C[n-1]*(4*n - 2)*(n+1)

代码取自kuangbin巨巨的模板

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;/** 完全大数模板* 输出cin>>a* 输出a.print();* 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。*/#define MAXN 9999#define MAXSIZE 1010#define DLEN 4class BigNum{private:    int a[500]; //可以控制大数的位数    int len;public:    BigNum() {        len=1;    //构造函数        memset(a,0,sizeof(a));    }    BigNum(const int); //将一个int类型的变量转化成大数    BigNum(const char*); //将一个字符串类型的变量转化为大数    BigNum(const BigNum &); //拷贝构造函数    BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算    friend istream& operator>>(istream&,BigNum&); //重载输入运算符    friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符    BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算    BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算    BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算    BigNum operator/(const int &)const; //重载除法运算符，大数对一个整数进行相除运算    BigNum 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 BigNum &T)const; //大数和另一个大数的大小比较    bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较    void print(); //输出大数};BigNum::BigNum(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;}BigNum::BigNum(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;    }}BigNum::BigNum(const BigNum &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];}BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算{    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,BigNum &b){    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,BigNum& b) //重载输出运算符{    int i;    cout<<b.a[b.len-1];    for(i=b.len-2; i>=0; i--) {        printf("%04d",b.a[i]);    }    return out;}BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算{    BigNum t(*this);    int i,big;    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;}BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算{    int i,j,big;    bool flag;    BigNum 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[len-1]==0 && t1.len>1) {        t1.len--;        big--;    }    if(flag)        t1.a[big-1]=0-t1.a[big-1];    return t1;}BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘{    BigNum 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;}BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算{    BigNum 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 BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模{    int i,d=0;    for(i=len-1; i>=0; i--)        d=((d*(MAXN+1))%b+a[i])%b;    return d;}BigNum BigNum::operator^(const int &n)const //大数的n次方运算{    BigNum 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 BigNum::operator>(const BigNum &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 BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较{    BigNum b(t);    return *this>b;}void BigNum::print() //输出大数{    int i;    printf("%d",a[len-1]);    for(i=len-2; i>=0; i--)        printf("%04d",a[i]);    printf("\n");}BigNum f[110];//卡特兰数int main(){    f[0]=1;    for(int i=1; i<=100; i++)        f[i]=f[i-1]*(4*i-2)/(i+1);//卡特兰数递推式    int n;    while(scanf("%d",&n)==1) {        if(n==-1)break;        f[n].print();    }    return 0;}

高精度，支持乘法和加法

#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; struct BigInt {     const static int mod = 10000;     const static int DLEN = 4;     int a[600], len;     BigInt()     {         memset(a, 0, sizeof a );         len = 1;     }     BigInt(int v)     {         memset(a, 0, sizeof a );         len = 0;         do {            a[len++] = v % mod;            v /= mod;         }while(v);     }     BigInt(const char s[])     {         memset(a, 0, sizeof a );         int L = strlen(s);         len = L/DLEN;         if(L%DLEN) len++;         int index = 0;         for(int i= L-1;i >= 0; i -= DLEN)         {             int t = 0;             int 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 operator + (const BigInt &b) const     {         BigInt res;         res.len = max(len,b.len);         for(int i = 0;i <= res.len;++i)            res.a[i] = 0;         for(int i = 0; i < res.len;++i)         {             res.a[i] += ((i < len)?a[i]:0) + ((i < b.len)?b.a[i]:0);             res.a[i+1] += res.a[i]/mod;             res.a[i] %= mod;         }         if(res.a[res.len] > 0) res.len++;         return res;     }     BigInt operator * (const BigInt &b)const     {         BigInt res;         for(int i = 0; i < len; ++i)         {             int up = 0;             for(int j = 0;j < b.len;++j)             {                 int temp = a[i]*b.a[j] + res.a[i+j] + up;                 res.a[i+j] = temp % mod;                 up = temp/mod;             }             if(up != 0)                res.a[i+b.len] = up;         }         res.len = len + b.len;         while(res.a[res.len - 1] == 0 && res.len > 1) res.len--;         return res;     }     void output()     {         printf("%d",a[len-1]);         for(int i = len-2; i >=0 ; i--)            printf("%04d",a[i]);         printf("\n");     } };int main(){    BigInt x, y;    char s1[1000],s2[1000];    scanf("%s%s",s1, s2);    x= BigInt(s1);    y = s2;    BigInt ans = x + y;    ans.output();    BigInt anssub = x * y;    anssub.output();    return 0;}