uscao Superprime Rib

没有想到什么好的方法，觉得dfs+米勒罗宾素性测试应该差不多，不过我还是开了一亿的素数表，然后..........打表.............

code：

/*ID: yueqiqLANG: C++TASK: sprime*/#include <set>#include <map>#include <ctime>#include <queue>#include <cmath>#include <stack>#include <limits>#include <vector>#include <bitset>#include <string>#include <cstdio>#include <cstring>#include <fstream>#include <string.h>#include <iostream>#include <algorithm>#define ls rt<<1#define rs rt<<1|1#define Si set<int>#define LL long long#define pb push_back#define PS printf(" ")#define Vi vector<int>#define LN printf("\n")#define SD(a) scanf("%d",&a)#define PD(a) printf("%d",a)#define SET(a,b) memset(a,b,sizeof(a))#define FF(i,a) for(int i(0);i<(a);i++)#define FD(i,a) for(int i(a);i>=(1);i--)#define FOR(i,a,b) for(int i(a);i<=(b);i++)#define FOD(i,a,b) for(int i(a);i>=(b);i--)#define readf freopen("sprime.in","r",stdin)#define writef freopen("sprime.out","w",stdout)const double pi = acos(-1.0);const int maxn = 1001;const int BigP = 99999999;const int INF = 99999999;const int dx[]={0,1,0,-1};const int dy[]={1,0,-1,0};using namespace std;int a[9][20];int main(){    readf;    writef;    a[1][0]=4;    a[1][1]=2;    a[1][2]=3;    a[1][3]=5;    a[1][4]=7;    a[2][0]=9;    a[2][1]=23;    a[2][2]=29;    a[2][3]=31;    a[2][4]=37;    a[2][5]=53;    a[2][6]=59;    a[2][7]=71;    a[2][8]=73;    a[2][9]=79;    a[3][0]=14;    a[3][1]=233;    a[3][2]=239;    a[3][3]=293;    a[3][4]=311;    a[3][5]=313;    a[3][6]=317;    a[3][7]=373;    a[3][8]=379;    a[3][9]=593;    a[3][10]=599;    a[3][11]=719;    a[3][12]=733;    a[3][13]=739;    a[3][14]=797;    a[4][0]=16;    a[4][1]=2333;    a[4][2]=2339;    a[4][3]=2393;    a[4][4]=2399;    a[4][5]=2939;    a[4][6]=3119;    a[4][7]=3137;    a[4][8]=3733;    a[4][9]=3739;    a[4][10]=3793;    a[4][11]=3797;    a[4][12]=5939;    a[4][13]=7193;    a[4][14]=7331;    a[4][15]=7333;    a[4][16]=7393;    a[5][0]=15;    a[5][1]=23333;    a[5][2]=23339;    a[5][3]=23399;    a[5][4]=23993;    a[5][5]=29399;    a[5][6]=31193;    a[5][7]=31379;    a[5][8]=37337;    a[5][9]=37339;    a[5][10]=37397;    a[5][11]=59393;    a[5][12]=59399;    a[5][13]=71933;    a[5][14]=73331;    a[5][15]=73939;    a[6][0]=12;    a[6][1]=233993;    a[6][2]=239933;    a[6][3]=293999;    a[6][4]=373379;    a[6][5]=373393;    a[6][6]=593933;    a[6][7]=593993;    a[6][8]=719333;    a[6][9]=739391;    a[6][10]=739393;    a[6][11]=739397;    a[6][12]=739399;    a[7][0]=8;    a[7][1]=2339933;    a[7][2]=2399333;    a[7][3]=2939999;    a[7][4]=3733799;    a[7][5]=5939333;    a[7][6]=7393913;    a[7][7]=7393931;    a[7][8]=7393933;    a[8][0]=5;    a[8][1]=23399339;    a[8][2]=29399999;    a[8][3]=37337999;    a[8][4]=59393339;    a[8][5]=73939133;    int N;    SD(N);    FOR(i,1,a[N][0]){        PD(a[N][i]);LN;    }    return 0;}