USACO1.5 Superprime Rib(sprime)

        求出一位的質數，存進vector_1中，再分別利用vector_1中的數做爲首位數求兩位質數，存入vector_2，依次求出位數爲n的質數，存進vector_n，打印時只需打印vector_n中的數字便可。

 

/*
ID:jzzlee1
PROG:sprime
LANG:C++
*/
#include<iostream>
#include<fstream>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
ifstream fin("sprime.in");
ofstream fout("sprime.out");
bool check(int p)		//檢測是否質數
{
	for(int i=2;i<=(int)sqrtf(p);i++)
		if(p%i==0)
			return 0;
	return 1;
}
int main()
{
	int n;
	fin>>n;
	//cin>>n;
	vector<int> vec1,vec2,vec3,vec4,vec5,vec6,vec7,vec8;
	int i=0;int d;int p;
	vec1.push_back(2);
	vec1.push_back(3);
	vec1.push_back(5);
	vec1.push_back(7);
	vector<int>::iterator iter;
	if(n>1)						//兩位數
	{
		for(iter=vec1.begin();iter!=vec1.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec2.push_back(p);
			}
	}
	if(n>2)						//三位數
	{
		for(iter=vec2.begin();iter!=vec2.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec3.push_back(p);
			}
	}
	if(n>3)						//四位數
	{
		for(iter=vec3.begin();iter!=vec3.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec4.push_back(p);
			}
	}
	if(n>4)						//五位數
	{
		for(iter=vec4.begin();iter!=vec4.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec5.push_back(p);
			}
	}
	if(n>5)						//六位數
	{
		for(iter=vec5.begin();iter!=vec5.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec6.push_back(p);
			}
	}
	if(n>6)						//七位數
	{
		for(iter=vec6.begin();iter!=vec6.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec7.push_back(p);
			}
	}
	if(n>7)						//八位數
	{
		for(iter=vec7.begin();iter!=vec7.end();++iter)
			for(d=1;d<=9;d+=2)
			{
				p=10*(*iter)+d;
				if(check(p))
					vec8.push_back(p);
			}
	}
	switch(n)
	{
		case 1:
			{
				for(iter=vec1.begin();iter!=vec1.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
		case 2:
			{
				for(iter=vec2.begin();iter!=vec2.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 3:
			{
				for(iter=vec3.begin();iter!=vec3.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 4:
			{
				for(iter=vec4.begin();iter!=vec4.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 5:
			{
				for(iter=vec5.begin();iter!=vec5.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 6:
			{
				for(iter=vec6.begin();iter!=vec6.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 7:
			{
				for(iter=vec7.begin();iter!=vec7.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
			case 8:
			{
				for(iter=vec8.begin();iter!=vec8.end();++iter)
					fout<<*iter<<endl;
					//cout<<*iter<<endl;
				break;
			}
	}
	return 0;
}