HDU 1016

xiaoxiao2021-02-28  12

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.  Note: the number of first circle should always be 1.    Input n (0 < n < 20).  Output The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.  You are to write a program that completes above process.  Print a blank line after each case.  Sample Input 6 8 Sample Output Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2

附代码

#include <iostream> #include <cstring> using namespace std; int n,a[25],book[25]; bool isPrime(int x) { if(x <= 1) return 0; for(int i = 2;i*i<=x;i++) if(x % i == 0) return 0; return 1; } void dfs(int x) { if(x == n + 1) { for(int i = 1;i<=n;i++) { if(isPrime(a[1] + a[n])) { if(i == n) cout<<a[n]<<endl; else cout<<a[i]<<" "; } } return; } else { for(int i = 2;i<=n;i++) { if(book[i] == 0) { if(isPrime(a[x-1] + i)) { a[x] = i; book[i] = 1; dfs(x+1); book[i] = 0; } else continue; } } } } int main() { int cnt = 1; while(cin >> n) { memset(a,0,sizeof(a)); memset(book,0,sizeof(book)); a[1] = 1; cout<<"Case "<<cnt++<<":"<<endl; dfs(2); cout<<endl; } }

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