Round and Round We Go
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10514 | Accepted: 4812 |
Description
A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a"cycle"of the digits of the original number. That is, if you consider the number after the last digit to "wrap around"back to the first digit, the sequence of digits in both numbers will be the same, though they may start at different positions.For example, the number 142857 is cyclic, as illustrated by the following table: 142857 *1 = 142857 142857 *2 = 285714 142857 *3 = 428571 142857 *4 = 571428
142857 *5 = 714285 142857 *6 = 857142
142857 *5 = 714285 142857 *6 = 857142
Input
Write a program which will determine whether or not numbers are cyclic. The input file is a list of integers from 2 to 60 digits in length. (Note that preceding zeros should not be removed, they are considered part of the number and count in determining n. Thus, "01"is a two-digit number, distinct from "1" which is a one-digit number.)
Output
For each input integer, write a line in the output indicating whether or not it is cyclic.
Sample Input
142857 142856 142858 01 0588235294117647
Sample Output
142857 is cyclic 142856 is not cyclic 142858 is not cyclic 01 is not cyclic 0588235294117647 is cyclic
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int N = 65;
char str[N];
int num[N],ans[N],len;
int cmp(const void *a,const void *b)
{
return *(int *)a-*(int *)b;
}
bool is_match()
{
qsort(num,len,sizeof(int),cmp);
qsort(ans,len,sizeof(int),cmp);
int i;
for(i=0;i<len;i++)
if(num[i]!=ans[i])
return false;
return true;
}
int main()
{
int i,j,k;
memset(num,0,sizeof(num));
memset(str,0,sizeof(str));
while(~scanf("%s",str))
{
len=strlen(str);
j=0;
memset(num,0,sizeof(num));
for(i=len-1;i>=0;i--)
num[j++]=str[i]-'0';
for(i=2;i<=len;i++)
{
memset(ans,0,sizeof(ans));
for(j=0;j<len;j++)
{
ans[j] = num[j]*i;
}
/*
for(k=0;k<len;k++)
printf("%d ",ans[k]);
putchar('\n');
*/
for(k=0;k<len;k++)//虽然最多增加2位 ， 但匹配函数是默认长度相同的
{
if(ans[k]>9)
{
ans[k+1] += ans[k]/10;//两条语句不可交换，搞了好久才发现
ans[k] %= 10;
}
}
bool flag = is_match();
if(flag)
{
if(i==len)
{
for(j=0;j<len;j++)
printf("%c",str[j]);
printf(" is cyclic\n");
memset(str,0,sizeof(str));
}
k=0;
for(j=len-1;j>=0;j--)
num[k++]=str[j]-'0';
}
else
{
for(j=0;j<len;j++)
printf("%c",str[j]);
printf(" is not cyclic\n");
memset(str,0,sizeof(str));
break;
}
}
}
return 0;
}
对于数N，若N为循环的则有N*(length(N)+1)=99....99， (length(N)个9)，length(N)为N的位数，含前导0。应该还是简单的...
//此为转载
//此为转载
#include<iostream>
#include<string>
using
namespace
std;
bool
fun(string str)
{
int
n=str.length()+1;
int
i,up=0,temp=0;
for
(i=n-2;i>=0;i--)
{
temp=(
int
)(str[i]-
'0'
);
if
((temp*n+up)%10!=9)
return
false
;
up=(temp*n+up)/10;
}
return
true
;
}
int
main()
{
string str1;
while
(cin>>str1)
{
if
(fun(str1))
{
cout<<str1<<
" is cyclic"
<<endl;
}
else
{
cout<<str1<<
" is not cyclic"
<<endl;
}
}
return
0;
}
