标签：include 一个 搜索 anti note alt 结果 tput mem
Inputn (0 < n < 20).
OutputThe 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
小范围数据，当然是递归+打表啦～这里掌握某种规律后也可以提高搜索效率，比如
素数环:给定n，1~n组成一个素数环，相邻两个数的和为素数。
首先偶数(2例外，但是本题不会出现两个数的和为2)不是素数，
所以素数环里奇偶间隔。如果n是奇数，必定有两个奇数相邻的情况。
所以当n为奇数时，输出“No Answer”。
当n == 1时只1个数，算作自环，输出1
所有n为偶数的情况都能变成奇偶间隔的环-----所以都有结果。
#include<stdio.h> #include<string.h> int n,c=0,i; int a[25],b[25]; int jo(int a) { return a%2==0?0:1; } int prime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0}; void dfs(int step) { int i; if(step>n&&prime[a[1]+a[n]]){ for(i=1;i<=n;++i){ if(i==1) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); return; } if(jo(a[step-1])){ for(i=2;i<=n;i+=2){ if(!b[i]&&prime[a[step-1]+i]){ b[i]=1; a[step]=i; dfs(step+1); b[i]=0; } } } else{ for(i=3;i<=n;i+=2){ if(!b[i]&&prime[a[step-1]+i]){ b[i]=1; a[step]=i; dfs(step+1); b[i]=0; } } } } int main() { while(~scanf("%d",&n)){ memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); a[1]=1;b[1]=1; if(n==1) printf("Case %d:\n1\n\n",++c); else if(!jo(n)){ printf("Case %d:\n",++c); dfs(2); printf("\n"); } } return 0; }
标签：include 一个 搜索 anti note alt 结果 tput mem
原文地址：http://www.cnblogs.com/yzm10/p/7230924.html
