九度1459：Prime ring problem

最新推荐文章于 2019-05-09 11:13:31 发布

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最新推荐文章于 2019-05-09 11:13:31 发布

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题目1459：Prime ring problem

时间限制：2 秒

内存限制：128 兆

特殊判题：否

提交：3027

解决：1239

题目描述：

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.

输入：

n (1 < n < 17).

输出：

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.

样例输入：

6
8

样例输出：

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

提示：

用printf打印输出。

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递归。本渣渣不会，借鉴的别人的代码，好好研读吧~

第一遍读不懂再读第二遍，第二遍都不懂再读第三遍.....

#include<cstdio>
#include<cmath>
#include<algorithm>
#define maxSize 10010
using namespace  std;

int cir[maxSize] = { 0 }; //记录排列
bool hashArray[maxSize] = { false };//是否加入队列

void myPrint(int n) {//打印
    for (int i = 1; i < n; i++) {
        printf("%d ", cir[i]);
    }
    printf("%d\n", cir[n]);
}

bool isPrime(int n) {//判断是否为素数
    int sqr = round(sqrt(n));
    for (int i = 2; i <= sqr; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}

void F(int n, int index) {
    if (index == n + 1) {//递归基
        //判断首尾相加是否为素数，若成功 打印
        if (isPrime(cir[n] + cir[1])) {
            myPrint(n);
        }
    }
    else {//0~index-1  已满足
        for (int i = 2; i <= n; i++) {
            if (hashArray[i] || (!isPrime(i + cir[index-1])) )
                continue;
            else {
                hashArray[i] = true;
                cir[index] = i;
                F(n, index + 1);//递归
                hashArray[i] = false;//恢复
            }
        }
    }
}

int main() {
    int n, num = 1;

    while (scanf("%d", &n) != EOF) {
        printf("Case %d:\n", num++);//输出编号
        cir[1] = 1;
        hashArray[1] = true;//起点
        F(n, 2);
        printf("\n");
    }
    return 0;
}

代码转自real_3stone

http://blog.youkuaiyun.com/qq_26398495/article/details/55803066

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坚持，而不是打鸡血~