PTA 1139 1138 1137 1136

PAT 1139 1138 1137 1136

一个月不写题，有点生疏。。脑子跟不上手速，还可以啦，反正今天很开心。

PAT 1139 First Contact

18/30 找个时间再修bug
23/30
28/30
30/30 补完啦

这道题的细节坑点:
	1.输出id需要补全4位  用print("%04d")  5分ok
	2.需要舍弃查询的相恋男女的直接边 3分ok
	3.大坑 0000 与 -0000 使用int型无法区分男女的. 2分ok   想法:用string输入 看长度和第一位符号位 判0和-0 转int 

思路：图论，找男女相恋的中间人。

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

/*
细节坑点:
	1.输出id需要补全4位  用print("%04d")  5分ok
	2.需要舍弃查询的相恋男女的直接边 3分ok
	3.大坑 0000 与 -0000 使用int型无法区分男女的. 2分ok   想法:用string输入 看长度和第一位符号位 判0和-0 转int 
*/ 

const int maxn = 10000;
vector<int> g[maxn];
int gender[10010];
int n,m;
vector<pair<int,int> >ans;

void init(){
	for(int i=0;i<=10000;i++) gender[i] = 0;
}

bool cmp(pair<int,int> a,pair<int,int> b){
	if(a.first == b.first) return a.second < b.second;
	return a.first < b.first;
}

int stringToAbsInt(string s){
	int x = 0;
	int flag = 0;
	if(s.length() == 5){
		flag = 1;
		for(int i=1;i<s.length();i++) x = x * 10 + (s[i]-'0');
	}else{
		flag = 0;
		for(int i=0;i<s.length();i++) x = x * 10 + (s[i]-'0');
	}
	if(flag){
		gender[x] = 1;
	}else{
		gender[x] = 0;
	}
	return abs(x);
}

int main(){
	cin>>n>>m;
	init();
	for(int i=1;i<=m;i++){
//		int u,v;
		string uu,vv;
		cin>>uu>>vv;
		int absu = stringToAbsInt(uu);
		int absv = stringToAbsInt(vv);
		g[absu].push_back(absv);
		g[absv].push_back(absu);
	}
	int k;
	cin>>k;
	for(int i=1;i<=k;i++){
		int u,v;
		cin>>u>>v;
		int absu = abs(u);
		int absv = abs(v);
		int genderu = gender[absu];
		int genderv = gender[absv];
		int cnt = 0;
		for(int c=0;c<g[absu].size();c++){
			int cc = g[absu][c];
			if(cc == absv) continue; //舍弃 v是u的直接朋友的边 
			if(gender[cc] != genderu ) continue;
			for(int d=0;d<g[absv].size();d++){
				int dd = g[absv][d];
				if(dd == absu) continue; //舍弃 u是v的直接朋友的边
				if(gender[dd] != genderv ) continue;  
				bool flag = false;
				for(int pos = 0;pos<g[cc].size();pos++){
					if(g[cc][pos] == dd){
						flag = true;
						break;
					}
				}
				if(flag){
					cnt++;
					ans.push_back(make_pair(g[absu][c],g[absv][d]));
				}
			}
		}
		cout<<cnt<<endl;
		sort(ans.begin(),ans.end(),cmp);
		for(int i=0;i<cnt;i++){
			printf("%04d %04d
",ans[i].first,ans[i].second); //输出补全4位 
		}
		ans.clear();
	}
	return 0;
}

1138 Postorder Traversal

数据结构，二叉树先序中序建树后后序遍历

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

const int maxn = 50010;
int n;
int pre[maxn];
int in[maxn];
int post[maxn];
bool flag = false;

struct node{
	int v;
	struct node *l;
	struct node *r;
};

//建树
node* build(int l,int r,int rootIdx){
	if(l > r) return NULL;
	node *root = new node();
	root->v = pre[rootIdx];
	int pos = l;
	while(in[pos] != pre[rootIdx]) pos++;
	root->l = build(l,pos-1,rootIdx+1);
	root->r = build(pos+1,r,rootIdx+(pos-l+1));
	return root;
}

//后序遍历 打印第一个被遍历的 结点值
void postTravel(node *Root){
	if(flag) return;
	if(Root->l == NULL && Root->r == NULL){
		cout<<Root->v<<endl;
		flag = true;
		return;
	}
	if(Root->l) postTravel(Root->l);
	if(Root->r) postTravel(Root->r);
	return;
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>pre[i];
	for(int i=1;i<=n;i++) cin>>in[i];
	node *Root = new node();
	Root = build(1,n,1); //建树
	postTravel(Root);
	return 0;
}

1137 Final Grading

数据结构，map，排序常见题型

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

const int maxn = 10010;
int p; //P the number of students having done the online programming assignmen
int m; //the number of students on the mid-term list
int n; //the number of students on the final exam list

struct student{
	string id;
	int gp;
	int gmid;
	int gfinal;
	int g;
};

unordered_map<string,student > mp;
vector<student> ans;

bool cmp(student stua,student stub){
	if(stua.g == stub.g){
		return stua.id < stub.id;
	}
	return stua.g > stub.g;
}

void print(){
	for(int i=0;i<ans.size();i++){
		cout<<ans[i].id<<" "<<ans[i].gp<<" "<<ans[i].gmid<<" "<<ans[i].gfinal<<" "<<ans[i].g<<endl;
	}
}

void input(){
	cin>>p>>m>>n;
	for(int i=1;i<=p;i++){
		string id;
		int gp;
		cin>>id>>gp;
		mp[id].id = id;
		mp[id].gp = gp;
		mp[id].gmid = -1;
		mp[id].gfinal = -1;
		mp[id].g = -1;
	}
	for(int i=1;i<=m;i++){
		string id;
		int gmid;
		cin>>id>>gmid;
		mp[id].gmid = gmid;
	}
	for(int i=1;i<=n;i++){
		string id;
		int gfinal;
		cin>>id>>gfinal;
		mp[id].gfinal = gfinal;
	}
}

int main(){
	input(); 
	for(auto it = mp.begin();it!=mp.end();it++){
		if(it->second.gp < 200) continue;
//		if(it->second.gfinal < 60 || it->second.gfinal > 100) continue; //吼吼 一开始题意理解错啦 g才是final score 
		if(it->second.gmid > it->second.gfinal) {
			int tempScore = ((it->second.gmid*4 + it->second.gfinal*6) );
			if(tempScore % 10 >= 5) it->second.g = tempScore/10+1;
			else it->second.g= tempScore/10;	
		}
		else it->second.g = it->second.gfinal;
		if(it->second.g >= 60 && it->second.g <=100) ans.push_back(it->second);
	}
	sort(ans.begin(),ans.end(),cmp);
	print();
	return 0;
} 

1136 A Delayed Palindrome

大数模拟加法进位

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

vector<int> v1;
vector<int> v2;
vector<int> ans;
/*
	reverse
	add
*/

void print(){
	for(int i=0;i<v1.size();i++) cout<<v1[i];
	cout<<" + ";
	for(int i=0;i<v2.size();i++) cout<<v2[i];
	cout<<" = ";
	for(int i=0;i<ans.size();i++) cout<<ans[i];
	cout<<endl; 
} 

bool check(){
	int len = ans.size();
	for(int i=0;i<len/2;i++){
		if(ans[i] != ans[len-i-1]) return false;
	}
	return true;
}

void clear(){
	v1.clear();
	v2.clear();
	for(int i=0;i<ans.size();i++) v1.push_back(ans[i]);
	ans.clear();
}
	
int main(){
	int d;
	while(scanf("%1d",&d)!=EOF) v1.push_back(d);
	int t = 0;
	bool flag = false;
	while((t++) < 10){
		for(int i=0;i<v1.size();i++) ans.push_back(v1[i]);
		if(check()){
			for(int i=0;i<ans.size();i++) cout<<ans[i];
			cout<<" is a palindromic number.";
			flag = true;
			break;
		}
		ans.clear();
		int ci = 0;
		for(int i=0;i<v1.size();i++) v2.push_back(v1[i]);
		reverse(v2.begin(),v2.end());
        //大数加法模拟
		for(int i=v1.size()-1;i>=0;i--){
			int si = v1[i] + v2[i] + ci;
			if(si >= 10) {
				ci = si/10; 
				si = si%10;
			}else{
				ci = 0;
			}
			ans.push_back(si);
		}
		if(ci!=0) ans.push_back(ci);
		reverse(ans.begin(),ans.end());//反转才是 结果和
		ci = 0;
		print();
		if(check()){
			for(int i=0;i<ans.size();i++) cout<<ans[i];
			cout<<" is a palindromic number.";
			flag = true;
			break;
		}
		clear();
	}
	if(flag == false) cout<<"Not found in 10 iterations."<<endl;
	return 0;
}