Round and Round We Go POJ 1047
2012-04-11 15:43
417 查看
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
Sample Output
就是没有找到为什么WA。气愤之下 就用了一位仁兄的打表法,果然好使,郁闷的是,他那个表中数据我的程序都过了。。。
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
做这个题的时候想法很简单,最笨的办法,直接遍历,但是不知道为什么WA了,求探讨。同时在和人探讨的过程中,也学到了一个新的解法:
#include "stdio.h" #include "String.h" #include "malloc.h" #define nMax 61 int judge(char*str,int i); void mutiply(char * str,int k,char * newstr); int roundJudge(char *a,char *b); void main() { char str[nMax]; int i,j,k,p,q; scanf("%s",str); i=strlen(str); if(judge(str,i)) printf("%s is cyclic",str); else printf("%s is not cyclic",str); } int judge(char*str,int i) { int j; char newstr[nMax]; for(j=2;j<=i;j++) { mutiply(str,j,newstr); if(!roundJudge(str,newstr)) { //printf("%d\n",j); return 0; break; } } return 1; } void mutiply(char * str,int k,char * newstr) { int i,j,p,q,*m; char kk[3]; if (k>=10) { kk[0]=k/10+'0'; kk[1]=k%10+'0'; kk[2]='\0'; } else { kk[0]=k+'0'; kk[1]='\0'; } i=strlen(str); j=strlen(kk); m=(int *)malloc(sizeof(int)*(i+j)); for(p=0;p<(i+j);p++) m[p]=0; for (p=0;p<i;p++) for(q=0;q<j;q++) m[p+q+1]+=(str[p]-'0')*(kk[q]-'0'); for(p=i+j-1;p>=0;p--) if(m[p]>=10) { m[p-1]+=m[p]/10; m[p]=m[p]%10; } p=0; while(m[p]==0) p++; for(q=0;p<i+j;q++,p++) newstr[q]=m[p]+'0'; newstr[q]='\0'; j=strlen(newstr); if(j<i) newstr[i]='\0'; for (p=i-1;p>=0;p--) { if(p>=(i-j)) newstr[p]=newstr[p-i+j]; else newstr[p]=0+'0'; } free(m); } int roundJudge(char *a,char *b) { int ca,cb,i,j,k; char *c,*temp; ca=strlen(a); cb=strlen(b); if(ca!=cb) return 0; else { c=(char*)malloc(sizeof(char)*(2*ca+1)); temp=(char *)malloc(sizeof(char)*(ca+1)); for (i=0;i<cb;i++) { c[i]=b[i]; } for(i=0;i<cb;i++) { c[strlen(b)+i]=b[i]; } c[2*ca]='\0'; for (i=0;i<ca;i++) { for (j=0;j<ca;j++) { temp[j]=c[i+j]; } temp[ca]='\0'; if(strcmp(a,temp)==0) { free(c); free(temp); return 1; break; } } free(c); free(temp); if(i==ca) return 0; } }
就是没有找到为什么WA。气愤之下 就用了一位仁兄的打表法,果然好使,郁闷的是,他那个表中数据我的程序都过了。。。
#include<iostream> #include<string> #include<map> using namespace std; map<string, int> dic; string sline; void init(){ string str[8] = {"142857","0588235294117647","052631578947368421","0434782608695652173913","0344827586206896551724137931", "0212765957446808510638297872340425531914893617","0169491525423728813559322033898305084745762711864406779661", "016393442622950819672131147540983606557377049180327868852459"}; for(int i=0;i<8;i++) dic[str[i]]=1; } int main(){ init(); while(cin>>sline){ if(dic.count(sline)==1) cout<<sline<<" is cyclic"<<endl; else cout<<sline<<" is not cyclic"<<endl; } }
相关文章推荐
- poj 1047 Round and Round We Go
- POJ-1047 Round and Round We Go
- POJ 1047 Round and Round We Go
- poj 1047-Round and Round We Go
- poj1047 Round and Round We Go
- POJ 1047 Round and Round We Go
- POJ 1047 Round and Round We Go
- POJ 1047 Round and Round We Go
- POJ 1047 Round and Round We Go (大数乘法) 水
- POJ 1047 Round and Round We Go
- POJ1047-Round and Round We Go
- POJ 1047 Round and Round We Go
- POJ 1047 Round and Round We Go
- poj 1047 Round and Round We Go
- OpenJudge百炼-2952-循环数 & poj-1047-Round and Round We Go-C语言-高精度计算
- POJ 1047 Round and Round We GO
- POJ 1047 Round and Round We Go
- poj-1047 Round and Round We Go
- poj 1047 Round and Round We Go
- poj 1047 Round And Round We Go