[2016-02-19][UVA][524][Prime Ring Problem]

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[2016-02-19][UVA][524][Prime Ring Problem]
UVA - 524
Prime Ring Problem
Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

 Status

Description

技术分享

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 技术分享 into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

技术分享


Note: the number of first circle should always be 1.

Input 

n (0 < n <= 16)

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.


You are to write a program that completes above process.

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


Miguel A. Revilla
1999-01-11
  • 时间:2016-02-19 21:16:47 星期五
  • 题目编号:UVA 524    
  • 题目大意:给定数字n,输出1~n的数字组成的排列,使得这些排列相邻的数字之和是素数(首位相连)
  • 方法:dfs枚举答案,回溯优化.
  • 解题过程遇到问题:
    • 格式问题:最后一组数据后面没有空行,否则WA
    • 每行最后一个数字后面没有空格,否则PE  



#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;
#define CLR(x,y) memset((x),(y),sizeof((x)))
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);(x)--)
#define FOR2(x,y,z) for((x)=(y);(x)<(z);(x)++)
#define FORD2(x,y,z) for((x)=(y);(x)>=(z);(x)--)
const int maxn = 20;
int ispri[maxn*2],n,vis[maxn],res[maxn];
void setpri(){
        ispri[1] = 1;
        CLR(ispri,-1);
        FOR(i,2,maxn*2){
                if(ispri[i]){
                        for(int j = i + i;j < maxn*2;j += i)
                                ispri[j] = 0;
                }
        }
}
void dfs(int cur){
        if(cur == n ){
                if(!ispri[res[0] + res[n - 1]]) return ;
                printf("%d",res[0]);
                FOR(i,1,n){
                        printf(" %d",res[i]);
                }
                puts("");
                return ;
        }
        FOR(i,2,n+1){
                if(!vis[i] && ispri[ res[cur - 1] + i ]){
                        vis[i] = 1;
                        res[cur] = i;
                        dfs(cur + 1);
                        vis[i] = 0;
                }
        }
}
int main(){
        int cntcase = 0;
        setpri();
        while(~scanf("%d",&n)){
                if(cntcase)     puts("");
                printf("Case %d:\n",++cntcase);
                CLR(vis,0);
                vis[1] = res[0] = 1;
                dfs(1);
        }
    return 0;
}  


























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