Day2-M-Prime Ring Problem-HDU1016

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.

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

思路：依旧是回溯法，逐个判断即可，注意首尾也需要判断一下，用筛法建个表，代码如下：

const int maxm = ;

int n, vis[maxm], prime[], res[maxm], kase = ;

void buildTable() {

    for (int i = ; i * i < ; ++i) {

        if(!prime[i]) {

            for (int j = i * i; j < ; j += i)

                prime[j] = ;

        }

    }

}

void dfs(int i) {

    if(i == n && !prime[res[] + res[n-]]) {

        for (int j = ; j < n; ++j) {

            if(j)printf(" ");

            printf("%d", res[j]);

        }

        printf("\n");

    }

    for (int j = ; j <= n; ++j) {

        if(!vis[j] && !prime[res[i-] + j]) {

            vis[j] = ;

            res[i] = j;

            dfs(i + );

            vis[j] = ;

        }

    }

}

int main() {

    buildTable();

    while(scanf("%d",&n) != EOF) {

        printf("Case %d:\n", ++kase);

        memset(vis, , sizeof(vis));

        res[] = ;

        vis[] = ;

        dfs();

        printf("\n");

    }

    return ;

}

巴特西

Day2-M-Prime Ring Problem-HDU1016

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