PAT（甲级）2018年冬季考试

1152 Google Recruitment

思路：判断素数

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1100;
int a[maxn];
int n,k;

long long getNum(int pos){
	long long x = 0;
	for(int i=pos;i<=pos+k-1;i++){
		x = x*10 + a[i];
	}
	return x;
}

bool prime(long long x){
	if(x < 2) return false;
	if(x == 2) return true;
	int endss = sqrt(x);
	for(int i=2;i<=endss;i++){
		if(x%i == 0) return false;
	}
	return true;
}

void print(long long  x){
	int len = 0;
	long long temp = x;
	while(temp){
		temp/=10;
		len++;
	}
	if(len == k) {
		printf("%lld",x);
		return;
	}
	int t = k - len;
	while(t--)printf("0");
	printf("%lld",x);
}

int main(){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++) 
		scanf("%1d",&a[i]);
	bool flag = false;
	for(int i=1;i+k-1<=n;i++){
		long long x = getNum(i);
		if(prime(x)){
			flag = true;
			print(x);
			break;
		}
	}
	if(flag == false) puts("404");
	return 0;
} 

1153 Decode Registration Card of PAT

思路：结构体排序，map计数，string.c_str()把字符串转成字符数组后可以使用print输出
注意：使用太多cin、cout太慢了会超时

#include<bits/stdc++.h>
//#include<unordered_map>
using namespace std;

/*
Cin和Cout不能用太多啦 
除了string 和 char类型的输入 最好把Cin替换成scanf
Cout全部换成printf 对于string类型的printf 需要将string转换成字符数组
string.c_str(); 
*/

const int maxn = 1e4+10;
int n,m;
struct node{
	char type;
	int site;
	string date;
	int num;
	int score;
	string all;
};
struct node stu[maxn];
vector<node> ans;

bool cmp1(node s1,node s2){
	if(s1.score == s2.score) {
		s1.all < s2.all;
	}
	return s1.score > s2.score;
}

bool cmp2(node s1,node s2){
	if(s1.score > s2.score) return true;
	else if(s1.score < s2.score) return false;
	return s1.all < s2.all;
}
int mm[1001];
struct nod{
	int site;
	int num;
	nod(int sss,int nnn){
		site = sss;
		num = nnn;
	}
};
vector<nod> ans3;
bool cmp3(nod s1,nod s2){
	if(s1.num == s2.num) return s1.site < s2.site;
	return s1.num > s2.num;
}
int main(){
//	cin>>n>>m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		string str;
		int score;
//		cin>>str>>score;
		cin>>str;
//		scanf("%s",&str);
		scanf("%d",&score);
		stu[i].all = str;
		stu[i].score = score;
		stu[i].type = str[0];
		stu[i].site = (str[1] - '0') * 100 + (str[2] - '0') * 10 + (str[3] - '0');
		stu[i].date = str.substr(4,6);
		stu[i].num = (str[10] - '0') * 100 + (str[11] - '0') * 10 + (str[12] - '0');
	}
	for(int Case=1;Case<=m;Case++){
		int Type;
//		cin>>Type;
		scanf("%d",&Type);
		int cnt = 0;
		ans.clear();
		ans3.clear();
		if(Type == 1){
			char type;
			cin>>type;
			printf("Case %d: 1 %c\n",Case,type);
			for(int i=1;i<=n;i++){
				if(stu[i].type == type){
					ans.push_back(stu[i]);
				}
			}
			cnt = ans.size();
			if(cnt == 0){
				puts("NA");
			}else{
				sort(ans.begin(),ans.end(),cmp2);
				for(int i=0;i<=cnt-1;i++){
//					cout<<ans[i].all<<" "<<ans[i].score<<endl;
					printf("%s ",ans[i].all.c_str());
					printf("%d\n",ans[i].score);
				}
			}
		}else if(Type == 2){
			int site;
//			cin>>site;
			scanf("%d",&site);
			printf("Case %d: 2 %d\n",Case,site);
			int tot = 0;
			for(int i=1;i<=n;i++){
				if(stu[i].site == site){
					tot += stu[i].score;
					cnt++;
				}
			}
			if(cnt == 0){
				puts("NA");
			}else{
				printf("%d %d\n",cnt,tot);
			}
		}else if(Type == 3){
			string date;
			cin>>date;
			printf("Case %d: 3 ",Case);
//			cout<<date<<endl;
			printf("%s\n",date.c_str());
			for(int i=100;i<=999;i++) mm[i] = 0;
			for(int i=1;i<=n;i++){
				if(stu[i].date == date){
					mm[stu[i].site]++;
				}
			}
			for(int i=100;i<=999;i++){
				if(mm[i] != 0){
					ans3.push_back(nod(i,mm[i]));
				}
			}
//			map<int, int> mp;
//			for(int i=1;i<=n;i++){
//				if(stu[i].date == date) mp[stu[i].site]++;
//			}
//			map<int, int>::iterator it = mp.begin(); 
//			while(it!=mp.end()){
//				ans3.push_back(nod(it->first,it->second));
//				it++;
//				cnt++;
//			}
			cnt = ans3.size();
			if(cnt == 0){
				puts("NA");
			}else{
				sort(ans3.begin(),ans3.end(),cmp3);		
				for(int i=0;i<=cnt-1;i++){
//					cout<<ans3[i].site<<" "<<ans3[i].num<<endl;
					printf("%d %d\n",ans3[i].site,ans3[i].num);
				}
			}
		}
	} 
	return 0;
} 

1154 Vertex Coloring

思路：图论，dfs遍历检查相邻结点的颜色

#include<bits/stdc++.h>
using namespace std;

const int maxn = 10010;
int n,m,k;
vector<int> g[maxn];
int col[maxn];
int vis[maxn];
set<int> se;
bool flag = false;

void dfs(int x){
	if(flag == false) return;
	vis[x] = 1;
	for(int i=0;i<g[x].size();i++){
		int v = g[x][i];
		if(col[x] == col[v]) flag = false;
		if(!vis[v]) dfs(v);
	}
}

int main(){
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	cin>>k;
	while(k--){
		se.clear();
		for(int i=0;i<n;i++){
			cin>>col[i];
			se.insert(col[i]);
			vis[i] = 0;
		}
		flag = true;
		for(int i=0;i<n;i++){
			if(flag == false) break;
			if(vis[i]) continue;
			dfs(i);
		}
		if(flag) cout<<se.size()<<"-coloring"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
} 

1155 Heap Paths

思路：给定序列，判断是否大根堆、小根堆(特殊二叉树)

#include<bits/stdc++.h>
using namespace std;

const int maxn = 50010;
int n;
int g[maxn];
vector<int> v;
bool flagMin = false;
bool flagMax = false;
bool flag = true;

void dfs(int x){
	v.push_back(g[x]);
	if(x * 2 + 1 <= n){
		dfs(x * 2 + 1);
	}
	if(x * 2 <= n){
		dfs(x * 2);
	}
	if(x * 2 + 1 > n && x * 2 > n){
		int temp = v[0];
		if(flag){
			int cnt = 0;
			for(int i=0;i<v.size();i++){
				if(i > 0 && v[i] >= v[i-1]) cnt++;
			}
			if(cnt == v.size() - 1) flagMax = true;
			else if(cnt == 0) flagMin = true;
			else flag = false;
		}
		for(int i=0;i<v.size();i++){
			if(i == 0) cout<<v[i];
			else cout<<" "<<v[i];
		}
		if(v.size() != 0) cout<<endl;
	}
	v.erase(v.end()-1);
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>g[i];
	dfs(1);
	if(flag == false){
		cout<<"Not Heap"<<endl;
	}else if(flagMin == true && flagMax == false){
		cout<<"Max Heap"<<endl;
	}else if(flagMin == false && flagMax == true){
		cout<<"Min Heap"<<endl;
	}else{
		cout<<"Not Heap"<<endl;
	}
	return 0;
}