poj 1426 数学推理找规律
2015-11-09 22:50
477 查看
Find The Multiple
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 22719 Accepted: 9362 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
2
6
19
0
Sample Output
10
100100100100100100
111111111111111111
题意:
输入一个不超过200的n,输出一个只有0和1组成的十进制数,使之为n的倍数
解法:
此题放到BFS里,可的确没看出来哪里用到宽搜了。 。我的做法是开一个较大的数组G[i],i代表这把这个只有0和1的十进制数看成是一个二进制数然后把这个二进制数转成十进制数,而G[i]即是这个只有0和1的十进制数对n取的余数,如当n = 6,则G[7] = 2(7 = 111,111%6 = 3),
当我们枚举生成所有情况的时候,可以看到每一个节点除了最后一个数字外(最后一个数字当节点数是奇数时为1,是偶数时为0),和父亲节点是一样的,那么,根据同余模定理可以有递推公式G[i] = (10*G[i/2] + i%2)%n.这样,当G[i]为0时可以判断找到这个数,然后将十进制的i变为二进制数输出即可。由此我觉得此题应该是个数学题。 。
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 22719 Accepted: 9362 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
2
6
19
0
Sample Output
10
100100100100100100
111111111111111111
题意:
输入一个不超过200的n,输出一个只有0和1组成的十进制数,使之为n的倍数
解法:
此题放到BFS里,可的确没看出来哪里用到宽搜了。 。我的做法是开一个较大的数组G[i],i代表这把这个只有0和1的十进制数看成是一个二进制数然后把这个二进制数转成十进制数,而G[i]即是这个只有0和1的十进制数对n取的余数,如当n = 6,则G[7] = 2(7 = 111,111%6 = 3),
当我们枚举生成所有情况的时候,可以看到每一个节点除了最后一个数字外(最后一个数字当节点数是奇数时为1,是偶数时为0),和父亲节点是一样的,那么,根据同余模定理可以有递推公式G[i] = (10*G[i/2] + i%2)%n.这样,当G[i]为0时可以判断找到这个数,然后将十进制的i变为二进制数输出即可。由此我觉得此题应该是个数学题。 。
#include <cstdio> #include <iostream> #include <stack> using namespace std; int G[50000000]; stack<int>st; int main(){ int n; while(scanf("%d",&n)&&n){ int i; G[1] = 1; for(i = 2;G[i-1];i++) G[i] = (10*G[i/2] + i%2)%n; i--; while(i){ int c = i&1; st.push(c); i >>= 1; } while(!st.empty()){ printf("%d",st.top()); st.pop(); } printf("\n"); } return 0; }