UVA - 10825 Anagram and Multiplication
2013-12-18 22:20
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题意:有m位n进制整数非常神奇:乘以2,3...,m之后,所有得到的数恰好是原数各数字的一个排列,输入m,n,求出这样的整数(最多只有一个)
思路:枚举0~n-1的数,因为乘以m次后得到不同的排列,所以我们可以根据这个求出m个数来,然后再用排列树搜索答案,最后check一下
思路:枚举0~n-1的数,因为乘以m次后得到不同的排列,所以我们可以根据这个求出m个数来,然后再用排列树搜索答案,最后check一下
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,m,a[10],p[10],b[10],cnt[405],vis[405]; int fk(){ int c[10]; for (int i = 0; i < m; i++) b[i] = c[i] = a[p[i]]; c[m] = 0; for (int k = 2; k <= m; k++){ for (int i = 0; i < m; i++){ c[i+1] += (c[i]+b[i])/n; c[i] = (c[i]+b[i]) % n; } if (c[m]) return 0; int flag = 1; for (int i = 0; i < m; i++) vis[c[i]]++; for (int i = 0; i < m; i++) if (vis[c[i]] != cnt[c[i]]) flag = 0; for (int i = 0; i < m; i++) vis[c[i]] = 0; if (!flag) return 0; } return 1; } int dfs(int u){ //排列树 if (u == m) return fk(); for (int i = u; i <= m-1; i++){ swap(p[i],p[u]); if (dfs(u+1)) return 1; swap(p[i],p[u]); } return 0; } int check(int x){ if (!x) return 0; int t = 0; for (int i = 0; i < m; i++){ t = (t + x) % n; a[i] = t; p[i] = i; } memset(cnt,0,sizeof(cnt)); memset(vis,0,sizeof(vis)); for (int i = 0; i < m; i++) cnt[a[i]]++; return dfs(0); } int main(){ while (scanf("%d%d",&m,&n) != EOF && n+m){ int flag = 0; for (int i = 0; i < n; i++) if (check(i)){ flag = 1; break; } if (!flag) printf("Not found.\n"); else { printf("%d",b[m-1]); for (int i = m-2; i >= 0; i--) printf(" %d",b[i]); printf("\n"); } } return 0; }
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