Zerojudge題解

一個放ZJ題解的地方

Easy

a001.哈囉

https://zerojudge.tw/ShowProblem?problemid=a001

#include <iostream>
using namespace std;

int main() {
string s;
 while(cin >> s){
  cout << "hello, "<< s << "\n";
 }
 return 0;
}

a002. 簡易加法

https://zerojudge.tw/ShowProblem?problemid=a002

#include<iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    cout<< a+b;
    return 0;
}

a003. 兩光法師占卜術

https://zerojudge.tw/ShowProblem?problemid=a003

#include <iostream>
using namespace std;

int main()
{
    int a, b;
    bool result=(a*2+b)%3;
    cin>>a>>b;
    if (result==0){
    cout<<"普通";
    }
    else if(result==1){
    cout<<"吉";
    }
    else if(result==2){
    cout<<"大吉";
    }
    return 0;
}

a004. 文文的求婚

https://zerojudge.tw/ShowProblem?problemid=a004

#include <iostream>
using namespace std;

int main() {
  int a;
	while(cin>>a){
    bool isleapa= (a % 4 == 0 && a % 100 != 0) || (a % 400 == 0);
    if(isleapa){
      cout<<"閏年\n";
     }
    else{
     cout<<"平年\n";
     }
    }
    return 0;
}

a005. Eva 的回家作業

https://zerojudge.tw/ShowProblem?problemid=a005

#include <iostream>
using namespace std;

int main() {
    int t;
    cin >> t; 

    for (int i = 0; i < t; i++) {
        int a, b, c, d; 
        
        cin >> a >> b >> c >> d;
        if (b - a == c - b && c - b == d - c) {
        
            int e = d + (d - c);
            cout << a << " " << b << " " << c << " " << d << " " << e <<'\n';
        }
        else if (b / a == c / b && c / b == d / c) {
        
            int e = d * (d / c);
            cout << a << " " << b << " " << c << " " << d << " " << e <<'\n';
        }
    }

    return 0;
}

a006. 一元二次方程式

https://zerojudge.tw/ShowProblem?problemid=a006

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    int d = b*b-4*a*c;
    
    if (d<0) {
    cout<<"No real root"<<'\n';
    }
    
    else if(d==0){
    double x1= -b/(2.0*a);
    cout<<"Two same roots x="<<x1<<'\n';
    }
    else{ 
    double x1= (-b+sqrt(d))/(2*a);
    double x2= (-b-sqrt(d))/(2*a);
          
    cout <<"Two different roots "<<"x1="<<x1<<" , x2="<<x2<<'\n';
    }
    return 0;
}

TOI

n630. 電影院 (Cinema)

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

struct movtime {
    int h;
    int m;
    int tm() const {
        return h * 60 + m;
    }
    void print() const {
    cout << setw(2) << setfill('0') << h << " "
            << setw(2) << setfill('0') << m << "\n";
    }
};

int main() {
    int n;
    cin >> n;
    vector<movtime> times(n);
    for (int i = 0; i < n; i++) {
        cin >> times[i].h >> times[i].m;
    }

    movtime now;
    cin >> now.h >> now.m;
    int ntm = now.tm() + 20;

    bool found = false;
    for (const movtime& movie : times) {
        if (movie.tm() >= ntm) {
            movie.print();
            found = true;
            break;
        }
    }

    if (!found) {
        cout << "Too Late" << endl;
    }

    return 0;
}

n631. 撲克 (Poker)

這題我一整個讀錯題義
我用觀察排序後的關係，去估算可湊出幾副牌和缺了多少張；這題應該是要用「每個牌號出現幾次」來精準對應牌組數量與缺少數。直接套用組合的數學關係即可。

錯誤寫法

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

int main(){
    int n,cnt=52,r=0;
    cin >> n;
    vector<int>poker(n);
    for(int i=0;i<n;i++){
        cin >> poker[i];
    }
    int d = n/52;
    for (int i = 1; i < poker.size(); i++) {
        if (poker[i] ==poker[i - 1]+1) cnt--;
    }
    sort(poker.begin(), poker.end());
    for (int i = 1; i < poker.size(); i++) {
        if (poker[i] == poker[i - 1]){
                r++;
                cnt+=52-r;
        }
    }
    cout << d << " " << cnt << "\n";
}

正解

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

int main(){
    int n;
    cin >> n;
    vector<int> cnt(53, 0);
    for(int i = 0; i < n; ++i){
        int k;
        cin >> k;
        cnt[k]++;
    }
    int last = *min_element(cnt.begin() + 1, cnt.end());
    int most = *max_element(cnt.begin() + 1, cnt.end());
    int tn = most * 52;
    int add = tn - n;

    cout << last << " " << add << "\n";
}