hdu1313 Round and Round We Go (大数乘法)
2016-02-21 21:37
393 查看
Problem Description
A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a ~{!0~}cycle~{!1~} of the digits of the original number. That is, if you consider the number after the last digit to ~{!0~}wrap around~{!1~} 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
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, ~{!0~}01~{!1~} is a two-digit number, distinct from ~{!0~}1~{!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
题意:给你一个字符串,让你判断它是不是一个“循环串”,循环串的定义是这个字符串所对应的整数乘上1~n(它的长度)的任何一个数,所得的结果为这个字符串循环后所得的整数。
思路:因为长度最多只有60,所以直接模拟就行了,附上大数乘法模板。
A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a ~{!0~}cycle~{!1~} of the digits of the original number. That is, if you consider the number after the last digit to ~{!0~}wrap around~{!1~} 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
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, ~{!0~}01~{!1~} is a two-digit number, distinct from ~{!0~}1~{!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
题意:给你一个字符串,让你判断它是不是一个“循环串”,循环串的定义是这个字符串所对应的整数乘上1~n(它的长度)的任何一个数,所得的结果为这个字符串循环后所得的整数。
思路:因为长度最多只有60,所以直接模拟就行了,附上大数乘法模板。
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> #include<vector> #include<map> #include<set> #include<queue> #include<stack> #include<string> #include<algorithm> using namespace std; typedef long long ll; typedef long double ldb; #define inf 99999999 #define pi acos(-1.0) #define eps 1e-15 #define maxn 200 #define Len 3000//大数的长度 using namespace std; int len; char a[Len],b[Len],c[Len]; void Mul(char a[],char b[],char c[])//大数乘法 { int i,j,alen,blen,clen; for(i=0; i<Len;i++){ c[i]='0'; } c[Len]='\0'; alen=strlen(a); blen=strlen(b); reverse(a,a+alen); reverse(b,b+blen); int sum=0; for(i=0; i<alen; i++){ for(j=0; j<blen; j++){ sum+=c[i+j]-'0'+(a[i]-'0')*(b[j]-'0'); c[i+j]=(sum%10)+'0'; sum/=10; } while(sum){ c[i+j++]+=sum%10; sum/=10; } } clen=len; c[clen]='\0'; reverse(c,c+clen); } char str[700],str1[700],str2[700]; struct node{ char s[70]; }d[70]; bool cmp(node a,node b){ return strcmp(a.s,b.s)<0; } int main() { int n,m,i,j,tot,alen,blen; while(scanf("%s",str1)!=EOF) { len=strlen(str1); for(i=0;i<len;i++){ str[i]=str1[i]; str[i+len]=str1[i]; } str[2*len]='\0'; for(i=1;i<=len;i++){ tot=i-1; for(j=0;j<len;j++){ d[i].s[j]=str[tot];tot++; } d[i].s[len]='\0'; } sort(d+1,d+1+len,cmp); int flag=1; for(i=2;i<=len;i++){ alen=len; for(j=0;j<len;j++){ a[j]=str[j]; } a[alen]='\0'; int tt=i; blen=0; while(tt){ b[blen++]=tt%10+'0'; tt/=10; } b[blen]='\0'; reverse(b,b+blen); Mul(a,b,c); if(strcmp(c,d[i].s)!=0){ flag=0; } } if(flag)printf("%s is cyclic\n",str1); else printf("%s is not cyclic\n",str1); } }
相关文章推荐
- 【HDU 3966】Aragorn's Story(未完待续)
- OC基础--分类(category) 和 协议(protocol)
- HDU 5556 Ancient Go 模拟
- HDU 5556 Ancient Go 模拟
- Golang学习笔记
- Golang学习笔记
- LeetCode 292. Nim Game
- Machine Learning - Regularized Linear Regression
- Google搜索
- django个人总结
- Machine Learning - Solving the Problem of Overfitting: Regularization
- django进行model字段的自定义
- go在windows上环境搭建
- HTC手机傻瓜式安装Google Play服务
- 544 - Heavy Cargo
- CodeForces 630 D. Hexagons!(水~)
- UVA_11090_GoingInCycle!!
- 最新google国内镜像站收集
- 78. Subsets
- ImportError: cannot import name gof