Google邀请题 数字跳跃 （大数+二分）

思路：

题意大体是这样，一个长度可达三百位的大数，每次可对该数加一，减一，除以二，问该数经过多少次操作可变为一，该数以字符串形式给出。就比如，15->16->8->4->2->1 ，15变为一需经过5次操作。一开始想着是用DFS做，但是好像有点小题大做，并且不是每次操作都会有三种选择，所以，本着使数能够尽量能除以2的原则，就使这个数不断靠近偶数，然后不断调整再除以2即可。因为是300位数，所以要用高精度大数（经过一晚上，对大数有了新的认识，找到了一个不错的模板）。

代码：

#include<iostream>#include<cstring>#include<iomanip>#include<algorithm>#include<cstdlib>#include<cstdio>using namespace std; #define MAXN 9999#define MAXSIZE 1000#define DLEN 4 class BigNum{public :    int a[MAXSIZE];    //大数的各个位数     int len;       //大数长度    BigNum(){ len = 1;memset(a,0,sizeof(a)); }      BigNum(const int);          BigNum(const char*);         BigNum(const BigNum &);      BigNum &operator=(const BigNum &);        friend istream& operator>>(istream&,  BigNum&);      friend ostream& operator<<(ostream&,  BigNum&);       BigNum operator+(const BigNum &) const;       BigNum operator-(const BigNum &) const;       BigNum operator*(const BigNum &) const;      BigNum operator/(const int   &) const;        BigNum operator^(const int  &) const;       int    operator%(const int  &) const;        bool   operator>(const BigNum & T)const;      bool   operator>(const int & t)const;         };BigNum::BigNum(const int b)     {    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)     {    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)  {    int i;    memset(a,0,sizeof(a));    for(i = 0 ; i < len ; i++)        a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum & n)   {    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; }BigNum BigNum::operator+(const BigNum & T) const  {    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   {    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   {    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   {    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    {    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   {    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    {    BigNum b(t);    return *this>b;}int main(void){    int i,n,ttt=1;     char s[300+5];    char a[2]="1";    int cnt=0;    scanf("%s",s);    BigNum z(s),y(a),num;    for(;;)    {    if(z%2==0)    {    z=z/2;    cnt++;    }    else if((z+y)/2%2==1)    {    z=z-y;    cnt++;    }    else if((z+y)/2%2==0)    {    z=z+y;    cnt++;    }    if(z.len==y.len)    {    if(z.a[z.len-1]==1)    {       printf("%d\n",cnt);        break;    }    }    }        return 0;}