hdu 4002 收获非常大的一个题 多功能大数模板的应用

Find the maximumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1277    Accepted Submission(s): 566


Problem DescriptionInputOutputSample InputSample OutputSourceRecommend 题意：输入cas 1 <=cas<=50000 情况数  输入N  从2<=n<=N中找到 n/φ(n)的最大值  2 ≤ N ≤ 10^100.  我一开始暴力打出 1-100000的表 找规律 发现有如下规律  ：n   i/phi（i） 的最大值1     22     2。。6    67    68    6。。29   630   3031  30。。209 30210 210211 210。。2310  23102311  2310。。可以看出 每一次变化都是各个质数的乘积2=26=2*330=2*3*5120=2*3*5*72310=2*3*5*7*11。。。 可以知道 每一次都是再上一次的基础上乘以下一个质数那么只要 把这些数找出来就可以了 对于输入n 输出正好大于等于n的第一个数但是问题来了那么大的数  没法乘出来啊  n可是10^100.这时候就要用大数了 另外要打表 打表才能过  否则超时对于下面的打表代码  简直就是极品啊  哈哈  所以一定要消化 以后肯定不少用    网上搜集思路    

#include <iostream>#include <cstring>#include<math.h>using namespace std;#define DIGIT4      //四位隔开,即万进制#define DEPTH10000        //万进制#define MAX     100typedef int bignum_t[MAX+1];/************************************************************************//* 读取操作数，对操作数进行处理存储在数组里                             *//************************************************************************/int read(bignum_t a,istream&is=cin){    char buf[MAX*DIGIT+1],ch ;    int i,j ;    memset((void*)a,0,sizeof(bignum_t));    if(!(is>>buf))return 0 ;    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for(i=1;i<=a[0];i++)    for(a[i]=0,j=0;j<DIGIT;j++)    a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ;    for(;!a[a[0]]&&a[0]>1;a[0]--);    return 1 ;}void write(const bignum_t a,ostream&os=cout){    int i,j ;    for(os<<a[i=a[0]],i--;i;i--)    for(j=DEPTH/10;j;j/=10)    os<<a[i]/j%10 ;}int comp(const bignum_t a,const bignum_t b){    int i ;    if(a[0]!=b[0])    return a[0]-b[0];    for(i=a[0];i;i--)    if(a[i]!=b[i])    return a[i]-b[i];    return 0 ;}int comp(const bignum_t a,const int b){    int c[12]=    {        1     }    ;    for(c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++);    return comp(a,c);}int comp(const bignum_t a,const int c,const int d,const bignum_t b){    int i,t=0,O=-DEPTH*2 ;    if(b[0]-a[0]<d&&c)    return 1 ;    for(i=b[0];i>d;i--)    {        t=t*DEPTH+a[i-d]*c-b[i];        if(t>0)return 1 ;        if(t<O)return 0 ;    }    for(i=d;i;i--)    {        t=t*DEPTH-b[i];        if(t>0)return 1 ;        if(t<O)return 0 ;    }    return t>0 ;}/************************************************************************//* 大数与大数相加                                                       *//************************************************************************/void add(bignum_t a,const bignum_t b){    int i ;    for(i=1;i<=b[0];i++)    if((a[i]+=b[i])>=DEPTH)    a[i]-=DEPTH,a[i+1]++;    if(b[0]>=a[0])    a[0]=b[0];    else     for(;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++);    a[0]+=(a[a[0]+1]>0);}/************************************************************************//* 大数与小数相加                                                       *//************************************************************************/void add(bignum_t a,const int b){    int i=1 ;    for(a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++);    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);}/************************************************************************//* 大数相减(被减数>=减数)                                               *//************************************************************************/void sub(bignum_t a,const bignum_t b){    int i ;    for(i=1;i<=b[0];i++)    if((a[i]-=b[i])<0)    a[i+1]--,a[i]+=DEPTH ;    for(;a[i]<0;a[i]+=DEPTH,i++,a[i]--);    for(;!a[a[0]]&&a[0]>1;a[0]--);}/************************************************************************//* 大数减去小数(被减数>=减数)                                           *//************************************************************************/void sub(bignum_t a,const int b){    int i=1 ;    for(a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void sub(bignum_t a,const bignum_t b,const int c,const int d){    int i,O=b[0]+d ;    for(i=1+d;i<=O;i++)    if((a[i]-=b[i-d]*c)<0)    a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ;    for(;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}/************************************************************************//* 大数相乘，读入被乘数a，乘数b，结果保存在c[]                          *//************************************************************************/void mul(bignum_t c,const bignum_t a,const bignum_t b){    int i,j ;    memset((void*)c,0,sizeof(bignum_t));    for(c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++)    for(j=1;j<=b[0];j++)    if((c[i+j-1]+=a[i]*b[j])>=DEPTH)    c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ;    for(c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--);}/************************************************************************//* 大数乘以小数，读入被乘数a，乘数b，结果保存在被乘数                   *//************************************************************************/void mul(bignum_t a,const int b){    int i ;    for(a[1]*=b,i=2;i<=a[0];i++)    {        a[i]*=b ;        if(a[i-1]>=DEPTH)        a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ;    }    for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);    for(;!a[a[0]]&&a[0]>1;a[0]--);}void mul(bignum_t b,const bignum_t a,const int c,const int d){    int i ;    memset((void*)b,0,sizeof(bignum_t));    for(b[0]=a[0]+d,i=d+1;i<=b[0];i++)    if((b[i]+=a[i-d]*c)>=DEPTH)    b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ;    for(;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH);    for(;!b[b[0]]&&b[0]>1;b[0]--);}/**************************************************************************//* 大数相除,读入被除数a，除数b，结果保存在c[]数组                         *//* 需要comp()函数                                                         *//**************************************************************************/void div(bignum_t c,bignum_t a,const bignum_t b){    int h,l,m,i ;    memset((void*)c,0,sizeof(bignum_t));    c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ;    for(i=c[0];i;sub(a,b,c[i]=m,i-1),i--)    for(h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1)    if(comp(b,m,i-1,a))h=m-1 ;    else l=m ;    for(;!c[c[0]]&&c[0]>1;c[0]--);    c[0]=c[0]>1?c[0]:1 ;}void div(bignum_t a,const int b,int&c){    int i ;    for(c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);    for(;!a[a[0]]&&a[0]>1;a[0]--);}/************************************************************************//* 大数平方根，读入大数a，结果保存在b[]数组里                           *//* 需要comp()函数                                                       *//************************************************************************/void sqrt(bignum_t b,bignum_t a){    int h,l,m,i ;    memset((void*)b,0,sizeof(bignum_t));    for(i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--)    for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1)    if(comp(b,m,i-1,a))h=m-1 ;    else l=m ;    for(;!b[b[0]]&&b[0]>1;b[0]--);    for(i=1;i<=b[0];b[i++]>>=1);}/************************************************************************//* 返回大数的长度                                                       *//************************************************************************/int length(const bignum_t a){    int t,ret ;    for(ret=(a[0]-1)*DIGIT,t=a[a[0]];t;t/=10,ret++);    return ret>0?ret:1 ;}/************************************************************************//* 返回指定位置的数字，从低位开始数到第b位，返回b位上的数               *//************************************************************************/int digit(const bignum_t a,const int b){    int i,ret ;    for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--);    return ret%10 ;}/************************************************************************//* 返回大数末尾0的个数                                                  *//************************************************************************/int zeronum(const bignum_t a){    int ret,t ;    for(ret=0;!a[ret+1];ret++);    for(t=a[ret+1],ret*=DIGIT;!(t%10);t/=10,ret++);    return ret ;}void comp(int*a,const int l,const int h,const int d){    int i,j,t ;    for(i=l;i<=h;i++)    for(t=i,j=2;t>1;j++)    while(!(t%j))    a[j]+=d,t/=j ;}void convert(int*a,const int h,bignum_t b){    int i,j,t=1 ;    memset(b,0,sizeof(bignum_t));    for(b[0]=b[1]=1,i=2;i<=h;i++)    if(a[i])    for(j=a[i];j;t*=i,j--)    if(t*i>DEPTH)    mul(b,t),t=1 ;    mul(b,t);}/************************************************************************//* 组合数                                                               *//************************************************************************/void combination(bignum_t a,int m,int n){    int*t=new int[m+1];    memset((void*)t,0,sizeof(int)*(m+1));    comp(t,n+1,m,1);    comp(t,2,m-n,-1);    convert(t,m,a);    delete[]t ;}/************************************************************************//* 排列数                                                               *//************************************************************************/void permutation(bignum_t a,int m,int n){    int i,t=1 ;    memset(a,0,sizeof(bignum_t));    a[0]=a[1]=1 ;    for(i=m-n+1;i<=m;t*=i++)    if(t*i>DEPTH)    mul(a,t),t=1 ;    mul(a,t);}#define SGN(x) ((x)>0?1:((x)<0?-1:0))#define ABS(x) ((x)>0?(x):-(x))int read(bignum_t a,int&sgn,istream&is=cin){    char str[MAX*DIGIT+2],ch,*buf ;    int i,j ;    memset((void*)a,0,sizeof(bignum_t));    if(!(is>>str))return 0 ;    buf=str,sgn=1 ;    if(*buf=='-')sgn=-1,buf++;    for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)    ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ;    for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for(i=1;i<=a[0];i++)    for(a[i]=0,j=0;j<DIGIT;j++)    a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ;    for(;!a[a[0]]&&a[0]>1;a[0]--);    if(a[0]==1&&!a[1])sgn=0 ;    return 1 ;}struct bignum {    bignum_t num ;    int sgn ;    public :    inline bignum()    {        memset(num,0,sizeof(bignum_t));        num[0]=1 ;        sgn=0 ;    }    inline int operator!()    {        return num[0]==1&&!num[1];    }    inline bignum&operator=(const bignum&a)    {        memcpy(num,a.num,sizeof(bignum_t));        sgn=a.sgn ;        return*this ;    }    inline bignum&operator=(const int a)    {        memset(num,0,sizeof(bignum_t));        num[0]=1 ;        sgn=SGN (a);        add(num,sgn*a);        return*this ;    }    ;    inline bignum&operator+=(const bignum&a)    {        if(sgn==a.sgn)add(num,a.num);        else if                 (sgn&&a.sgn)        {            int ret=comp(num,a.num);            if(ret>0)sub(num,a.num);            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memcpy(num,a.num,sizeof(bignum_t));                sub (num,t);                sgn=a.sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if(!sgn)memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ;        return*this ;    }    inline bignum&operator+=(const int a)    {        if(sgn*a>0)add(num,ABS(a));        else if(sgn&&a)        {            int  ret=comp(num,ABS(a));            if(ret>0)sub(num,ABS(a));            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memset(num,0,sizeof(bignum_t));                num[0]=1 ;                add(num,ABS (a));                sgn=-sgn ;                sub(num,t);            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if (!sgn)sgn=SGN(a),add(num,ABS(a));        return*this ;    }    inline bignum operator+(const bignum&a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        ret.sgn=sgn ;        ret+=a ;        return ret ;    }    inline bignum operator+(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        ret.sgn=sgn ;        ret+=a ;        return ret ;    }    inline bignum&operator-=(const bignum&a)    {        if(sgn*a.sgn<0)add(num,a.num);        else if                 (sgn&&a.sgn)        {            int ret=comp(num,a.num);            if(ret>0)sub(num,a.num);            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memcpy(num,a.num,sizeof(bignum_t));                sub(num,t);                sgn=-sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if(!sgn)add (num,a.num),sgn=-a.sgn ;        return*this ;    }    inline bignum&operator-=(const int a)    {        if(sgn*a<0)add(num,ABS(a));        else if(sgn&&a)        {            int  ret=comp(num,ABS(a));            if(ret>0)sub(num,ABS(a));            else if(ret<0)            {                bignum_t t ;                memcpy(t,num,sizeof(bignum_t));                memset(num,0,sizeof(bignum_t));                num[0]=1 ;                add(num,ABS(a));                sub(num,t);                sgn=-sgn ;            }            else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ;        }        else if (!sgn)sgn=-SGN(a),add(num,ABS(a));        return*this ;    }    inline bignum operator-(const bignum&a)    {        bignum ret ;        memcpy(ret.num,num,sizeof(bignum_t));        ret.sgn=sgn ;        ret-=a ;        return ret ;    }    inline bignum operator-(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof(bignum_t));        ret.sgn=sgn ;        ret-=a ;        return ret ;    }    inline bignum&operator*=(const bignum&a)    {        bignum_t t ;        mul(t,num,a.num);        memcpy(num,t,sizeof(bignum_t));        sgn*=a.sgn ;        return*this ;    }    inline bignum&operator*=(const int a)    {        mul(num,ABS(a));        sgn*=SGN(a);        return*this ;    }    inline bignum operator*(const bignum&a)    {        bignum ret ;        mul(ret.num,num,a.num);        ret.sgn=sgn*a.sgn ;        return ret ;    }    inline bignum operator*(const int a)    {        bignum ret ;        memcpy(ret.num,num,sizeof (bignum_t));        mul(ret.num,ABS(a));        ret.sgn=sgn*SGN(a);        return ret ;    }    inline bignum&operator/=(const bignum&a)    {        bignum_t t ;        div(t,num,a.num);        memcpy (num,t,sizeof(bignum_t));        sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ;        return*this ;    }    inline bignum&operator/=(const int a)    {        int t ;        div(num,ABS(a),t);        sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a);        return*this ;    }    inline bignum operator/(const bignum&a)    {        bignum ret ;        bignum_t t ;        memcpy(t,num,sizeof(bignum_t));        div(ret.num,t,a.num);        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ;        return ret ;    }    inline bignum operator/(const int a)    {        bignum ret ;        int t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(ret.num,ABS(a),t);        ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a);        return ret ;    }    inline bignum&operator%=(const bignum&a)    {        bignum_t t ;        div(t,num,a.num);        if(num[0]==1&&!num[1])sgn=0 ;        return*this ;    }    inline int operator%=(const int a)    {        int t ;        div(num,ABS(a),t);        memset(num,0,sizeof (bignum_t));        num[0]=1 ;        add(num,t);        return t ;    }    inline bignum operator%(const bignum&a)    {        bignum ret ;        bignum_t t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(t,ret.num,a.num);        ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ;        return ret ;    }    inline int operator%(const int a)    {        bignum ret ;        int t ;        memcpy(ret.num,num,sizeof(bignum_t));        div(ret.num,ABS(a),t);        memset(ret.num,0,sizeof(bignum_t));        ret.num[0]=1 ;        add(ret.num,t);        return t ;    }    inline bignum&operator++()    {        *this+=1 ;        return*this ;    }    inline bignum&operator--()    {        *this-=1 ;        return*this ;    }    ;    inline int operator>(const bignum&a)    {        return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0);    }    inline int operator>(const int a)    {        return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0);    }    inline int operator>=(const bignum&a)    {        return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0);    }    inline int operator>=(const int a)    {        return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0);    }    inline int operator<(const bignum&a)    {        return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0);    }    inline int operator<(const int a)    {        return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0);    }    inline int operator<=(const bignum&a)    {        return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0);    }    inline int operator<=(const int a)    {        return sgn<0?(a<0?comp(num,-a)>=0:1):        (sgn>0?(a>0?comp(num,a)<=0:0):a>=0);    }    inline int operator==(const bignum&a)    {        return(sgn==a.sgn)?!comp(num,a.num):0 ;    }    inline int operator==(const int a)    {        return(sgn*a>=0)?!comp(num,ABS(a)):0 ;    }    inline int operator!=(const bignum&a)    {        return(sgn==a.sgn)?comp(num,a.num):1 ;    }    inline int operator!=(const int a)    {        return(sgn*a>=0)?comp(num,ABS(a)):1 ;    }    inline int operator[](const int a)    {        return digit(num,a);    }    friend inline istream&operator>>(istream&is,bignum&a)    {        read(a.num,a.sgn,is);        return  is ;    }    friend inline ostream&operator<<(ostream&os,const bignum&a)    {        if(a.sgn<0)os<<'-' ;        write(a.num,os);        return os ;    }    friend inline bignum sqrt(const bignum&a)    {        bignum ret ;        bignum_t t ;        memcpy(t,a.num,sizeof(bignum_t));        sqrt(ret.num,t);        ret.sgn=ret.num[0]!=1||ret.num[1];        return ret ;    }    friend inline bignum sqrt(const bignum&a,bignum&b)    {        bignum ret ;        memcpy(b.num,a.num,sizeof(bignum_t));        sqrt(ret.num,b.num);        ret.sgn=ret.num[0]!=1||ret.num[1];        b.sgn=b.num[0]!=1||ret.num[1];        return ret ;    }    inline int length()    {        return :: length(num);    }    inline int zeronum()    {        return :: zeronum(num);    }    inline bignum C(const int m,const int n)    {        combination(num,m,n);        sgn=1 ;        return*this ;    }    inline bignum P(const int m,const int n)    {        permutation(num,m,n);        sgn=1 ;        return*this ;    }};/*int main(){   bignum a,b,c; cin>>a>>b;cout<<"加法:"<<a+b<<endl;cout<<"减法:"<<a-b<<endl;cout<<"乘法:"<<a*b<<endl;cout<<"除法:"<<a/b<<endl;c=sqrt(a);cout<<"平方根:"<<c<<endl;cout<<"a的长度:"<<a.length()<<endl;cout<<"a的末尾0个数:"<<a.zeronum()<<endl<<endl;cout<<"组合: 从10个不同元素取3个元素组合的所有可能性为"<<c.C(10,3)<<endl;cout<<"排列: 从10个不同元素取3个元素排列的所有可能性为"<<c.P(10,3)<<endl;    return 0 ;}*/  //////////////////////////////////////////////////////////////*上面是一个完整的大数模板  已经功能的演示   我只是在下面修改了主函数和加入了 get_prime */int vis[1000],c;  int prime[200];  void get_prime()  {      int i,j,n,m;      c=0;      n=1000;      m=(int)sqrt(n+0.5);      memset(vis,0,sizeof(vis));      for(i=2;i<=m;i++)           if(!vis[i])          {              for(j=i*i;j<=n;j+=i)                  vis[j]=1;          }      for(i=2;i<=n;i++) if(!vis[i])          prime[c++]=i;  }  int main(){   bignum a[60],b,n;int i;    get_prime();a[0]=2;  for(i=1;i<60&&i<c;i++)  {  cout<<a[i-1]<<endl;  b=prime[i];  a[i]=a[i-1]*b;    }    return 0 ;}

AC代码  

#include<stdio.h>#include<string.h>char cc[][500]=//打表，存的是前n个素数的乘积，节俭时候 {        "2",        "6",        "30",        "210",        "2310",        "30030",        "510510",        "9699690",        "223092870",        "6469693230",        "200560490130",        "7420738134810",        "304250263527210",        "13082761331670030",        "614889782588491410",        "32589158477190044730",        "1922760350154212639070",         "117288381359406970983270",        "7858321551080267055879090",        "557940830126698960967415390",        "40729680599249024150621323470",        "3217644767340672907899084554130",        "267064515689275851355624017992790",        "23768741896345550770650537601358310",        "2305567963945518424753102147331756070",        "232862364358497360900063316880507363070",        "23984823528925228172706521638692258396210",        "2566376117594999414479597815340071648394470",        "279734996817854936178276161872067809674997230",        "31610054640417607788145206291543662493274686990",        "4014476939333036189094441199026045136645885247730",        "525896479052627740771371797072411912900610967452630",        "72047817630210000485677936198920432067383702541010310",        "10014646650599190067509233131649940057366334653200433090",        "1492182350939279320058875736615841068547583863326864530410",        "225319534991831177328890236228992001350685163362356544091910",        "35375166993717494840635767087951744212057570647889977422429870",        "5766152219975951659023630035336134306565384015606066319856068810",        "962947420735983927056946215901134429196419130606213075415963491270",        "166589903787325219380851695350896256250980509594874862046961683989710",        "29819592777931214269172453467810429868925511217482600306406141434158090",        "5397346292805549782720214077673687806275517530364350655459511599582614290",        "1030893141925860008499560888835674370998623848299590975192766715520279329390",        "198962376391690981640415251545285153602734402721821058212203976095413910572270",        "39195588149163123383161804554421175259738677336198748467804183290796540382737190",        "7799922041683461553249199106329813876687996789903550945093032474868511536164700810",        "1645783550795210387735581011435590727981167322669649249414629852197255934130751870910",        "367009731827331916465034565550136732339800312955331782619462457039988073311157667212930",        "83311209124804345037562846379881038241134671040860314654617977748077292641632790457335110",        "19078266889580195013601891820992757757219839668357012055907516904309700014933909014729740190",        "4445236185272185438169240794291312557432222642727183809026451438704160103479600800432029464270",        "1062411448280052319722448549835623701226301211611796930357321893850294264731624591303255041960530",        "256041159035492609053110100510385311995538591998443060216114576417920917800321526504084465112487730",            "64266330917908644872330635228106713310880186591609208114244758680898150367880703152525200743234420230"};char str[110];int main(){    int T;    int i;    scanf("%d",&T);    while(T--)    {        scanf("%s",&str);        int x=strlen(str);        for(i=0;i<60;i++)        {            int y=strlen(cc[i]);            if(x<y) break;            if(x>y)continue;            if(strcmp(str,cc[i])<0) break;        }            printf("%s\n",cc[i-1]);    }        return 0;}