PAT(甲级)2019年春季考试

7-1 Sexy Primes 判断素数 一个点没过17/20分

错因:输出i-6写成了输出i,当时写的很乱,可以参考其他人的写法

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

const int maxn = 1e8;
typedef long long ll;
int n;

bool isprime(ll x){
	if(x == 0 || x==1 || x==2) return false;
	for(ll i=2;i<=sqrt(x);i++){
		if(x%i==0) return false;
	}
	return true;
}

int main(){
	scanf("%lld",&n);
	if(isprime(n)){
		if(n>=6 && isprime(n-6)){
			puts("Yes");
			printf("%lld",n-6);
		}else if(isprime(n+6)){
			puts("Yes");
			printf("%lld",n+6);
		}else{
			puts("No");
			for(ll i=n+1;i<=maxn;i++){
				if(i-6>=0){
					if((i-6)>=0 && isprime(i) && isprime(i-6)){
						printf("%lld",i);
						break;
					}
				}else{
					if(isprime(i) && isprime(i+6)){
						printf("%lld",i);
						break;
					}
				}
			}
		}
	}else{
		puts("No");
		for(ll i=n+1;i<=maxn;i++){
			if(i-6>=0){
				if((i-6)>=0 && isprime(i) && isprime(i-6)){
					printf("%lld",i);
					break;
				}
			}else{
				if(isprime(i) && isprime(i+6)){
					printf("%lld",i);
					break;
				}
			}
		}
	}
	return 0;
} 

修改后的代码:

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

const int maxn = 1e8+10;
typedef long long ll;
int n;

bool isprime(ll x){
	if(x == 0 || x==1 || x==2) return false;
	for(ll i=2;i<=sqrt(x);i++){
		if(x%i==0) return false;
	}
	return true;
}

int main(){
		scanf("%lld",&n);
		printf("i = %d\n",ac);
		if(isprime(n)){
			if(n>=6 && isprime(n-6)){
				puts("Yes");
				printf("%lld",n-6);
			}else if(isprime(n+6)){
				puts("Yes");
				printf("%lld",n+6);
			}else{
				puts("No");
				for(ll i=n+1;i<=maxn;i++){
					if(i-6>=0){
						if((i-6)>=0 && isprime(i) && isprime(i-6)){
							printf("%lld",i);
							break;
						}
					}else{
						if(isprime(i) && isprime(i+6)){
							printf("%lld",i);
							break;
						}
					}
				}
			}
		}else{
			puts("No");
			for(ll i=n+1;i<=maxn;i++){
				if(i-6>=0){
					if((i-6)>=0 && isprime(i) && isprime(i-6)){
						printf("%lld",i - 6);
						break;
					}else if(isprime(i) && isprime(i+6)){
						printf("%lld",i);
						break;
					}
				}else{
					if(isprime(i) && isprime(i+6)){
						printf("%lld",i);
						break;
					}
				}
			}
		}
	return 0;
} 

7-2 Anniversary 集合、查找排序 25分

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

set<string> se1;
set<string> se2;
int n,m;

int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		string id;
		cin>>id;
		se1.insert(id);
	}
	int ans = 0;
	string guestold = "99999999";
	string assold = "99999999";
	string guest;
	string ass;
	cin>>m;
	for(int i=1;i<=m;i++){
		string id;
		cin>>id;
		se2.insert(id);
		string temp;
		for(int j=6;j<=13;j++) temp.push_back(id[j]);
		if(se1.find(id) != se1.end()){
			ans++;
			if(assold > temp) {
				assold = temp;
				ass = id;
			}
		}
		if(guestold > temp) {
			guestold = temp;
			guest = id;
		}
	}
	cout<<ans<<endl;
	if(ans != 0){
		if(n !=0){
			cout<<ass<<endl;
		}
	}else{
		if(m!=0){
			cout<<guest<<endl;
		}
	}
	return 0;
}

7-3 Telefraud Detection 图论 15/25分

更新错误原因:没有完全搞清题意,total duration,表示两人之间累计的通话时间
下面的代码只算了最后1次时间(覆盖)
修改后,可以用并查集做,不过感觉没必要,统计入度出度+暴力判断集合也不会超时的
参考链接



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

/*
邻接矩阵
点:出度 和他的边相关的入度
*/

const int maxn = 1010;
int k,n,m;
int g[maxn][maxn];
vector<int> sss;
int vis[maxn];
vector<int> temp;
set<int> se;


void print(){
	if(temp.size()==0) return;
	cout<<temp[0];
	for(int i=1;i<temp.size();i++){
		cout<<" "<<temp[i];
	}
	cout<<endl;
}

int main(){
	cin>>k>>n>>m;
	for(int i=1;i<=m;i++){
		int u,v,w;
		cin>>u>>v>>w;
		g[u][v] = w;
	}
	for(int i=1;i<=n;i++){
		int call = 0;
		int back = 0;
		for(int j=1;j<=n;j++){
			if(g[i][j]!=0 && g[i][j] <= 5){
				call++;
				if(g[j][i] != 0) back++;
			}
		}
		if(call > k && back*5 <= call){
			sss.push_back(i);
			se.insert(i);
		}
	}
	if(sss.size() == 0){
		cout<<"None"<<endl;
		return 0;
	}
	sort(sss.begin(),sss.end());
	for(int i=0;i<sss.size();i++){
		int x = sss[i];
		temp.clear();
		if(!vis[x]){
			vis[x] = 1;
			temp.push_back(x);
			for(int j=i+1;j<sss.size();j++){
				//判断第sss[i]能否可以用 
				bool can = true;
				for(int k=0;k<temp.size();k++){
					if(g[sss[j]][temp[k]] == 0){
						can = false;
						break;
					}
				}
				if(can == true){
					vis[sss[j]] = 1;
					temp.push_back(sss[j]);
				}
			}
			print();
		}
	}
	return 0;
}

7-4 Structure of a Binary Tree 二叉树中序后序建树,dfs深搜30分

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

int n,m;
const int maxn = 1100;
int in[maxn];
int post[maxn];

struct node{
	int v;
	node* l;
	node* r;
};
int level[maxn];
int l[maxn];
int r[maxn];
int sib[maxn];
int par[maxn];
bool full = true;

void dfs(node *root,int depth){
	if(root == NULL) return;
	int x = root->v;
	level[x] = depth;
	if(root->l != NULL){
		par[root->l->v] = x;
		l[x] = root->l->v;
		dfs(root->l,depth+1);
	}
	if(root->r != NULL){
		par[root->r->v] = x;
		r[x] = root->r->v;
		dfs(root->r,depth+1);
	}
	if(root->l != NULL && root->r != NULL){
		sib[root->l->v] = root->r->v;
		sib[root->r->v] = root->l->v;
	}
	if((root->l == NULL && root->r !=NULL) || (root->r == NULL && root->l !=NULL) ){
		full = false;
	}
}

node* build(int il,int ir,int pl,int pr){
	if(il > ir) return NULL;
	int rootv = post[pr];
	int pos = il;
	while(pos <= ir && in[pos] != rootv) pos++;
	node *root = new node;
	root->v = rootv;
	root->l = build(il,pos-1,pl,pr-(ir-pos)-1);
//	pr-(ir-pos)+1
//	root->r = build(pos+1,ir,pl+(pos-il)+1,pr);
	root->r = build(pos+1,ir,pr-(ir-pos)+1,pr-1);
	return root;
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>post[i];
	for(int j=1;j<=n;j++) cin>>in[j];
	node* Root = build(1,n,1,n);
	dfs(Root,1);
	cin>>m;
	getchar();
	while(m--){
		string stat;
	    getline(cin, stat);
		if(stat.find("root") != string::npos){
			int root = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				root = root*10 + (stat[i] - '0');
			}
			if(root == Root->v){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("siblings") != string::npos){
			int parent = 0;
			int pos = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				parent = parent*10 + (stat[i] - '0');
				pos = i;
			}
			int child = 0;
			for(int i = pos+1;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') continue;
				child = child*10 + (stat[i] - '0');
			}
			if(sib[parent] == child && sib[child] == parent){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("parent") != string::npos){
			int parent = 0;
			int pos = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				parent = parent*10 + (stat[i] - '0');
				pos = i;
			}
			int child = 0;
			for(int i = pos+1;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') continue;
				child = child*10 + (stat[i] - '0');
			}
			if(par[child] == parent){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("left") != string::npos){
			int parent = 0;
			int pos = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				parent = parent*10 + (stat[i] - '0');
				pos = i;
			}
			int child = 0;
			for(int i = pos+1;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') continue;
				child = child*10 + (stat[i] - '0');
			}
			if(l[child] == parent){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("right") != string::npos){
			int parent = 0;
			int pos = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				parent = parent*10 + (stat[i] - '0');
				pos = i;
			}
			int child = 0;
			for(int i = pos+1;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') continue;
				child = child*10 + (stat[i] - '0');
			}
			if(r[child] == parent){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("same") != string::npos){
			int parent = 0;
			int pos = 0;
			for(int i=0;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') break;
				parent = parent*10 + (stat[i] - '0');
				pos = i;
			}
			int child = 0;
			for(int i = pos+1;i<stat.length();i++){
				if(stat[i] > '9' ||  stat[i] < '0') continue;
				child = child*10 + (stat[i] - '0');
			}
			if(level[parent] == level[child]){
				puts("Yes");
			}else{
				puts("No");
			}
		}else if(stat.find("full") != string::npos){
			if(full) puts("Yes");
			else puts("No");
		}
		
	}
	return 0;
}
posted @ 2019-09-24 18:10  fishers  阅读(1261)  评论(0编辑  收藏  举报