POJ 1426 Find The Multiple(数论——中国同余定理)
题目链接:
http://poj.org/problem?id=1426
Description
Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.
Input
The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.
Output
For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.
Sample Input
2 6 19 0
Sample Output
10 100100100100100100 111111111111111111
题意描述:
输入n (1 <= n <= 200)
输出一个由0和1组成的数并且能够被n整除
解题思路:
看了题解有了一些思路,用mod[i]=mod[i/2]*10+i%2;这个式子计算二进制i在十进制的形式并存进mod[i]数组,所以使用long long 定义mod数组,很容易理解
1 #include<stdio.h> 2 long long mod[600001]; 3 int main() 4 { 5 int n; 6 while(scanf("%d",&n),n) 7 { 8 int i; 9 for(i=1; mod[i-1]%n != 0;i++) 10 mod[i]=mod[i/2]*10+i%2;// mod[i]表示十进制形式的二进制i 11 printf("%I64d\n",mod[i]); 12 } 13 return 0; 14 }
但是延伸一下,如果超过200或者更大呢,请参考
#include<stdio.h> #define N 600000 int mod[N]; int ans[200];//根据情况增大数组 int main() { int i,k,n,j; while(scanf("%d",&n),n) { mod[1]=1%n; for(i=2;mod[i-1]!=0;i++) mod[i]=(mod[i/2]*10+i%2) %n; //printf("i-1=%d\n",i-1); i--; k=0; while(i) { ans[k++]=i%2; i/=2; } for(i=k-1;i>=0;i--) printf("%d",ans[i]); puts("\n"); } return 0; }
欢迎交流,共同进步——