大数运算

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本文介绍了一个使用C++实现的程序，该程序用于判断一个数是否为循环数。循环数是指一个整数，在乘以从1到其位数长度之间的任意整数时，其数字序列会形成一个循环。文章通过示例解释了循环数的概念，并提供了完整的源代码。

C++ 中一直错用cstring,今天终于出错了，改回string 才好了

Round and Round We Go

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10549		Accepted: 4830

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

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

代码如下
#include<iostream>
#include<cstdio>
#include<string>
using namespace std;

struct Node
{
int a[80];
int digit;
}num1,num2;
int sign1[80];

bool compare(int k,int sum ) //判断是否为一个颠倒顺序的数组
{
int i,j;
for(i=k,j=0;i<sum;i++,j++)
if(num1.a[i]!=num2.a[j])
return false;

for(i=0;i<k;i++,j++)
if(num1.a[i]!=num2.a[j])
return false;

return true;
}

int main()
{
int i,j,k;
int temp;
bool ans;
string str;
//while(scanf("%s",str)!=EOF)
while(cin>>str) //改为cstring时编译出错
{
num1.digit=str.length();
for(i=0,j=num1.digit-1;i<num1.digit;i++,j--)
{
num1.a[i]=str[j]-'0';
}
for(i=2;i<=num1.digit;i++)
{
int carry=0;
ans=false;
memset(sign1,0,sizeof(sign1));
for(j=0;j<num1.digit;j++)
{
temp=num1.a[j]*i+carry;
carry=temp/10;
num2.a[j]=temp%10;
}
if(carry)
//{printf("%s is not cyclic\n",str);break;}
{cout<<str<<" is not cyclic"<<endl; break;}
else
{
for(k=0;k<num1.digit;k++)
if(num1.a[k]==num2.a[0]&&sign1[k]==0)
{
sign1[k]=1;
ans=compare(k,num1.digit);
if(ans)
break; //防止数组里有相同的数而再判断为false
}
if(!ans)
//{printf("%s is not cyclic\n",str);break;;}
{cout<<str<<" is not cyclic"<<endl; break;}
}//end else

}//end for (31)
if(ans)
//printf("%s is cyclic\n",str);
cout<<str<<" is cyclic"<<endl;

}//end while
return 0;
}