POJ 1049 Round and Round We Go 大数模拟
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Round and Round We Go
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12611 Accepted: 5900
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 *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
142857142856142858010588235294117647
Sample Output
142857 is cyclic142856 is not cyclic142858 is not cyclic01 is not cyclic0588235294117647 is cyclic
Source
Greater New York 2001
题意:给出一个大数,位数为n,将其从一乘到n,若得到的每一个数都可以根据原数本身顺序得到即为cyclic。
分析: 直接大数模拟即可,具体见代码:
//大数 模拟得结果#include<stdio.h>#include<string.h>using namespace std;const int maxn=65;char s[maxn];int a[maxn],temp[maxn];int main(){while(~scanf("%s",s)){int n=strlen(s);for(int i=0; i<n; i++){a[i]=s[i]-'0';temp[i]=a[i];}int flag=1;for(int i=1; i<=n; i++)//i为每次的乘数{int f=1;for(int j=0; j<n; j++){a[j]=temp[j];a[j]*=i;}for(int j=n-1; j>=1; j--){a[j-1]+=a[j]/10;a[j]%=10;}if(a[0]>9)//发生了进位 所得结果位数大于原位数{flag=0;break;}f=0;int ff=0,s=-1;//ff判断是否能够匹配for(int j=0; j<n; j++)if(temp[j]==a[0])//需要注意的是 需要对temp中每个和起点相同的点进行匹配{s=j;int fff=1;for(int k=0; k<n; k++){if(a[k]!=temp[s]){fff=0;break;}else{s++;s%=(n);}}if(fff){ff=1;//能够有一种排列匹配break;}}if(s==-1)f=0;if(!ff)//不能够匹配{flag=0;break;}}if(flag)printf("%s is cyclic\n",s);elseprintf("%s is not cyclic\n",s);}return 0;}//对于其他的数num,如果其位数是n,如果num*(n+1)得到的结果是n个9,那么这个数就是可循环的。特记下,以备后日回顾。
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