hdu 5050 Divided Land（高精度）

唉。。当时板子找不到了，然后从网上找了个板子。。。结果板子有错。。减法的一个地方写错了。比赛完才找到。。简直坑

以后就用这个板子了。

这道题我比较蠢，是把2进制转成十进制再用大数的GCD，因为没有重载大数与大数的除法，所以不能用欧几里德。

有几个规律

1. 如果a,b是偶数那么gcd(a,b) = gcd(a/2,b/2)*2

2.如果a是偶数b是奇数，那么gcd(a,b) = gcd(a/2,b).

3.如果a,b都是奇数,且a>b那么gcd(a,b) = gcd((a-b)/2,b)

都很好证明就不说了。用这个办法可以很轻松的算出大数与大数的gcd，用二进制模拟也是一样的。时间复杂度是o(log(max(a,b)))级别的因为每步至少其中一个都除了一个2。

但是要注意不能写成递归形式不然会爆栈，手工扩栈可以解决～不过直接写循环很简单的～

AC代码：

#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>#include<ctime>#include<string.h>#include<string>#include<bitset>using namespace std;#define ll __int64#define eps 1e-10#define MOD 20071027#define MAXN 9999#define MAXSIZE 10#define DLEN 4template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}class BigNum{public:    int a[100];    //可以控制大数的位数    int len;       //大数长度    BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数    BigNum(const int);       //将一个int类型的变量转化为大数    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    BigNum operator>>(const int &) const;    BigNum operator<<(const int &) const;    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    bool   operator<(const BigNum & T)const;    bool   operator<(const int & t)const;    bool   operator==(const BigNum & T)const;    bool   operator==(const int & t)const;    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,  BigNum & 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,  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--)    {        cout.width(DLEN);        cout.fill('0');        cout << b.a[i];    }    return out;}BigNum BigNum::operator>>(const int &k) const{    BigNum t(*this);    int p = k;    while(p)    {        t = t/2;        p--;    }    return t;}BigNum BigNum::operator<<(const int &k) const{    BigNum t(*this);    int p = k;    while(p)    {        t = t*2;        p--;    }    return t;}BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算{    BigNum 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;}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[t1.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运算{    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运算{    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(ln >= 0 && a[ln] == T.a[ln])            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;}bool BigNum::operator == (const BigNum & T)const{    if(len!=T.len) return false;    int ln = len-1;    int flag = 0;    while(ln >= 0 && a[ln] == T.a[ln]) ln--;    if(ln == -1) return true;    return false;}bool BigNum::operator == (const int & t)const{    BigNum b(t);    return *this==b;}bool BigNum::operator < (const BigNum & T) const{    if(!(*this>T)&&!(*this==T)) return true;    return false;}bool BigNum::operator < (const int & t) const{    BigNum b(t);    return *this<b;}void BigNum::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;}bool iseven(BigNum x){    if(x.a[0]%2==0) return true;    return false;}char a[2005],b[2005];BigNum gcd(BigNum a,BigNum b){    BigNum k = 1,ans;    while(1)    {        if(a == 0)        {            ans = b;            break;        }        if(b == 0)        {            ans = a;            break;        }        if(a < b)        {            BigNum temp = a;            a = b;            b = temp;        }        else        {            if(iseven(a))            {                if(iseven(b))                {                    a = a / 2;                    b = b / 2;                    k=k*2;                }                else                {                    a = a/2;                }            }            else            {                if(iseven(b)) b = b/2;                else a = (a-b)/2;            }        }    }    return ans*k;}int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o4.txt","w",stdout);#endif // GLQ    int t,i,j,cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%s %s",a,b);        BigNum temp(1),t1(0),t2(0);        int len = strlen(a);        for(i = len-1; i >= 0; i--)        {            if(a[i] == '1') t1 = t1+temp;            temp = temp*2;        }        len = strlen(b);        temp = 1;        for(i = len-1; i >= 0; i--)        {            if(b[i] == '1') t2 = t2+temp;            temp = temp*2;        }        int tt[2005];//        t1 = "11498698008513698797986710471";//        t2 = "3733414046940010605264291687";        BigNum ans=gcd(t1,t2);        int k = 0;        while(!(ans==0))        {            if(ans%2==1)                tt[k++] = 1;            else tt[k++] = 0;            ans = ans/2;        }        printf("Case #%d: ",cas++);        for(i = k-1; i >= 0; i--)            printf("%d",tt[i]);        printf("\n");    }    return 0;}