ZOJ 1530 find the multiple 找倍数

 ZOJ Problem Set - 1530

Find The Multiple

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Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

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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 (0) 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

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Source: Asia 2002, Dhaka (Bengal)

此题DFS(BFS，看你的性格，不过我相信C coder选DFS，BFS的queue还得自己搞，C++直接模板。。。)，模拟一下高精度除法，这里在《C语言经典算法大全》里截个图

此题本人用了一个随机化才0ms，这是为了避免有“左倾”或者“右倾”——一味的搜索做字数或者右子树，左倾的话AC大概30ms。

随机化的判断开始用

if(rand()%2);//判断奇偶，不过时间代价太大，原因在于取模运算太慢

后来用

if(rand()/2>RAND_MAX);//除法也太慢

于是换成代码中的乘法

/******************************************** *problem name:zoj 1530 find the multiple * *coder:mike * *date:2011-02-05 * *email:creativewang@163.com * *note:nothing * *********************************************/ #include<stdio.h> #include<stdlib.h> #include<time.h> #include<string.h> #define MAX_LEN 99 int finish=0; void print(int*base,int len) { int i=0; while(i<len) printf("%d",base[i++]); putchar('/n'); return; } int dfs(int* lnum,int depth,int remainder,int n) { if(depth>MAX_LEN) return 0; int tmp=remainder*10; if(!(tmp+1)%n) { lnum[depth-1]=1; finish=1; print(lnum,depth); } if(!(tmp%n)) { lnum[depth-1]=0; print(lnum,depth); finish=1; } if(rand()*2>RAND_MAX) { if(finish) return 0; lnum[depth-1]=0; dfs(lnum,depth+1,tmp%n,n); if(finish) return 0; lnum[depth-1]=1; dfs(lnum,depth+1,(tmp+1)%n,n); } else { if(finish) return 0; lnum[depth-1]=1; dfs(lnum,depth+1,(tmp+1)%n,n); if(finish) return 0; lnum[depth-1]=0; dfs(lnum,depth+1,tmp%n,n); } return 0; } int main(void) { srand((unsigned)time(NULL)); int n; int num[MAX_LEN]; while(scanf("%d",&n)!=EOF) { if(!n) break; finish=0; memset(num,0,sizeof(int)*MAX_LEN); num[0]=1; dfs(num,2,1,n); } return 0; }

另一个傻傻的TLE版本

#include<stdio.h> #include<stdlib.h> #include<string.h> #define BASE 10 #define MAX_LEN 120 int hash[2][BASE]= /**< hash[m][n]；m->0-1串末尾数字；n->n末位数字*/ { {0,1,5,7,5,2,5,3,5,9}, {1,9,5,3,5,4,5,7,5,1} }; int add(int* lnum,int len,int addend) { int carry=0; int i; lnum[len-1]+=addend; for(i=len-1;i>=0;i--) { lnum[i]+=carry; carry=lnum[i]/10; lnum[i]%=10; if(carry==0) break; } return 0; } int check(int*lnum,int len) { while(len-->0) if(lnum[len]>1) return 0; return 1; } int print(int* lnum,int len) { int i=0; while(lnum[i]==0) i++; while(i<len) printf("%d",lnum[i++]); putchar('/n'); return 0; } int main(void) { int n; int num[MAX_LEN]; int last; while(scanf("%d",&n)!=EOF) { memset(num,0,MAX_LEN*sizeof(int)); last=n%10; if(n==0) break; do add(num,MAX_LEN,hash[num[MAX_LEN-1]][last]*n); while(!check(num,MAX_LEN)); print(num,MAX_LEN); } return 0; }