POJ 1426 Find The Multiple(BFS 同余模定理)
2015-04-01 21:34
465 查看
题意 给你一个数n 输出一个仅由0,1组成的数m使得m是n的倍数
找到一个m 是m%n==0 就行了 初始让m=1 然后bfs扩展m的位数 只有两种情况 m = m * 10 或 m = m*10 + 1;
同余模定理 (a+b) % c = (a%c + b%c) % c, (a*b)%c = (a%c * b%c) % c;
运用同余模定理 可以只记录余数 这样就可以避免处理大数了 因为余数不会超过n 而n是小于两百的
对于已经得到过的余数 是可以跳过的 可以考虑下为什么
于是只用保存每一位选的是0还是1 只要保证最后余数为0就行了
1. 选0 m = m*10 % n;
2. 选1 m = (m*10 + 1) % n;
Find The Multiple
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
Sample Output
找到一个m 是m%n==0 就行了 初始让m=1 然后bfs扩展m的位数 只有两种情况 m = m * 10 或 m = m*10 + 1;
同余模定理 (a+b) % c = (a%c + b%c) % c, (a*b)%c = (a%c * b%c) % c;
运用同余模定理 可以只记录余数 这样就可以避免处理大数了 因为余数不会超过n 而n是小于两百的
对于已经得到过的余数 是可以跳过的 可以考虑下为什么
于是只用保存每一位选的是0还是1 只要保证最后余数为0就行了
1. 选0 m = m*10 % n;
2. 选1 m = (m*10 + 1) % n;
#include <cstdio> using namespace std; const int N = 205; int q , p , d , n; int bfs() { int le = 0, ri = 0, r = 1, cr; int v = {0}; v[q[ri++] = r % n] = d[0] = 1; p[0] = -1; while(le < ri) { cr = q[le]; if(cr == 0) return le; if(!v[r = cr * 10 % n]) v[r] = 1, p[ri] = le, d[ri] = 0, q[ri++] = r; if(!v[r = (cr * 10 + 1) % n]) v[r] = 1, p[ri] = le, d[ri] = 1, q[ri++] = r; ++le; } return -1; } void print(int k) { if(p[k] >= 0) print(p[k]); printf("%d", d[k]); } int main() { while(scanf("%d", &n), n) { print(bfs()); puts(""); } return 0; }
Find The Multiple
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
相关文章推荐
- POJ1426-Find The Multiple (BFS 余数)
- poj 1426 Find The Multiple (bfs 搜索)
- POJ 1426 Find The Multiple(BFS)
- POJ1426 Find The Multiple(bfs)
- poj 1426 Find The Multiple(bfs)
- poj 1426 Find The Multiple( bfs )
- POJ-1426-Find The Multiple【BFS】
- poj 1426 Find The Multiple (BFS)
- POJ - 1426 Find The Multiple(15.10.10 搜索专题)bfs
- POJ_1426_Find The Multiple(BFS)
- poj - 1426-Find The Multiple-BFS
- poj 1426 Find The Multiple(bfs)
- POJ 1426-Find The Multiple(bfs)
- Poj 1426--Find The Multiple(bfs或dfs)
- POJ训练计划1426_Find The Multiple(BFS)
- Find The Multiple (poj 1426 bfs)
- POJ 1426 Find The Multiple(BFS 水~)
- POJ 1426 Find The Multiple --- BFS || DFS
- POJ 1426 Find The Multiple (bfs搜索)
- DFS/BFS(同余模) POJ 1426 Find The Multiple