程序代码：

#include<stdio.h>

int main(void)
{
    int i=2,j=0,k;
    double aa[10];
    double ff=0.0;
    while (scanf("%lf",&aa[j]))
    {
        if(aa[j]<=0.0000001) break;
        j++;
    }
    k=j;
    j=0;
    while(j<k)
    {
        if(aa[j]<0.5) {printf("%d card(s)\n",1);  j++;ff=0.0;i=2;continue;}
        ff+=(double)1.0/i;
        if(ff>=aa[j]) {printf("%d card(s)\n",i-1);j++;i=2;ff=0.0;continue;}
        i++;
    }

    return 0;
}

程序代码：

#include <stdio.h>
#define max 100
int i=0;
void dec2yyy(int yyy,int source,char* bin)
{
    if(source==0) return;
    dec2yyy(yyy,source/yyy,bin);
    if((source%yyy)<10) bin[i]=(char)(source%yyy+0x30);
    else bin[i]=(char)(source%yyy+0x37);
    i++;
}
int main(void)
{
    char s[max];
    int j=0,k,k1,k2,len=0;
    int flag=1;  
    int flag2=0; 
    int ss[20];
    char* begin=s;
    char* tmp=s;
    char* end=NULL;
    int pan[21]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
    for(j=0;j<max;j++) s[j]='\0';
    j=0;
    while (scanf("%d",&ss[j]))
    {
        if(ss[j]==0) break;
        j++;
    }
    k=j;
    int jj=0;
    int p=0;
    while(ss[jj]!=0)
    {
        for(k=2;k<=16;k++)
        {
            dec2yyy(k,ss[jj],s);    
            for(k1=0;k1<max;k1++)  
            {
                if(s[k1]=='\0')
                {
                    end=begin+len-1;
                    break;
                }
                len++;
            }
            if(begin==end){flag=1;goto _next;}
            while(begin!=end)  
            {
                if((end-begin==1) && (*begin==*end)) {flag=1;break;}  
                if(*begin!=*end)
                {
                    flag=0;
                    break;
                }
                else
                {
                    begin++;
                    end--;
                }
            }
_next:
            if(flag==1)   
            {
                pan[p++]=k;
                flag2++;
            }
            for(j=0;j<max;j++) {s[j]='\0';i=0;}  
            len=0;
            flag=1;
            begin=tmp;
        }
        if(flag2>0)
        {
            printf("Number %d is palindrom in basis",ss[jj]);
            k2=0;
            while(pan[k2]!=0){printf(" %d",pan[k2]);k2++;}
            printf("\n");
        }
        for(k2=0;k2<15;) pan[k2++]=0;
        if(flag2<1) printf("Number %d is not a palindrom\n",ss[jj]);
        flag2=0;
        jj++;
        p=0;
    }
    return 0;
}

程序代码：

F:\c_source\t2\Debug>fc
1
2
3
11
12
13
0
Number  1 is palindrom in basis 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number  2 is palindrom in basis 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number  3 is palindrom in basis 2 4 5 6 7 8 9 10 11 12 13 14 15 16
Number 11 is palindrom in basis 10 12 13 14 15 16
Number 12 is palindrom in basis 5 11 13 14 15 16
Number 13 is palindrom in basis 3 12 14 15 16

程序代码：

Statement of the Problem
We say that a number is a palindrom if it is the sane when read from left to right or from right to left. For example, the number 75457 is a palindrom.

Of course, the property depends on the basis in which is number is represented. The number 17 is not a palindrom in base 10, but its representation in base 2 (10001) is a palindrom.

The objective of this problem is to verify if a set of given numbers are palindroms in any basis from 2 to 16.



Input Format

Several integer numbers comprise the input. Each number 0 < n < 50000 is given in decimal basis in a separate line. The input ends with a zero.



Output Format

Your program must print the message Number i is palindrom in basis where I is the given number, followed by the basis where the representation of the number is a palindrom. If the number is not a palindrom in any basis between 2 and 16, your program must print the message Number i is not palindrom.



Sample Input

17
19
0



Sample Output

Number 17 is palindrom in basis 2 4 16
Number 19 is not a palindrom