Find The Multiple(poj 1426)
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Find The Multiple
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33035 Accepted: 13823 Special Judge
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
26190
Sample Output
10100100100100100100111111111111111111
Source
Dhaka 2002
这道题,好多人都是拿long long 就给过了,但是如果数据再大一点,用 long long 就过不了了。
首先我们得明白 每个数的二进制数 乘以 10;就等于他原来的十进制树乘以 2.
举个例子 : 1 二进制 1 如果 1 乘以 10等于10 那就是2的二进制,10*10=100 那就是4的二进制。那么一个十进制数除以2得到的数,他们在二进制里相差10倍。
还得明白 同余模定理
(a*b)%n = (a%n *b%n)%n;
(a+b)%n = (a%n +b%n)%n;
这些都懂了之后,这道题,就比较好理解了,我们主要就是枚举每一个数的二进制数,看看如果这是个十进制的,能不能是n的倍数。如果单纯的去求这样的二进制肯定会超出int 的范围。那么就需要同余模定理了。
#include<stdio.h>int mod_er[1000000]; //这里必须开得足够大,才能满足能够枚举到这样的shint main(){ int n; while(scanf("%d",&n)!=EOF) { int ans[200]; if(n==0) return 0; mod_er[1]=1; int i; for( i=2;mod_er[i-1]!=0;i++) { mod_er[i]=(mod_er[i/2]*10+i%2)%n; //一个个的枚举每个数的二进制,是否满足条件。 } i--; int k=0; while(i) { int temp=i%2; ans[k]=temp; k++; i=i/2; } for(int j=k-1;j>=0;j--) { printf("%d",ans[j]); } printf("\n"); }}
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