Contest3897 - 计科23级算法设计与分析上机作业-01

题目链接

A.判断一个数能否被3,5,7整除

题面

思路

直接模拟即可

示例代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'

const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);

    int n;
    cin >> n;
    if(n % 3 == 0 && n % 5 == 0 && n % 7 == 0) cout << "3 5 7" << '\n';
    else if(n % 3 == 0 && n % 5 == 0) cout << "3 5" << '\n';
    else if(n % 3 == 0 && n % 7 == 0) cout << "3 7" << '\n';
    else if(n % 5 == 0 && n % 7 == 0) cout << "5 7" << '\n';
    else if(n % 3 == 0) cout << "3" << '\n';
    else if(n % 5 == 0) cout << "5" << '\n';
    else if(n % 7 == 0) cout << "7" << '\n';
    else cout << 'n' << '\n';

    return 0;
}

B.判断是否构成回文

题面

思路

双指针直接模拟即可

示例代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'

const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);

    string s;
    cin >> s;
    int i = 0, j = s.size() - 2;
    while(i < j){
        if(s[i] != s[j]){cout << "No\n"; return 0;}
        i++;
        j--;
    }
    cout << "Yes\n";
    return 0;
}

C.求最大公约数

题面

思路

辗转相除法直接模拟即可

示例代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'

const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;

int gcd(int a, int b){
    return b ? gcd(b, a % b) : a;
}

void solve() {
    int a, b;
    cin >> a >> b;
    cout << gcd(a, b) << '\n';
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while(T--) solve();
    return 0;
}

D.HTML新手 - 图片收集者

题面

思路

用getline读取每一行字符串并转化成stringstream流，检测line的开头是否为<img，是的话后面就有一个图片地址，在后面再检测word前几位字符是否为src=，再提取图片地址，最后合并输出即可。但是不知道为啥WA了。

示例代码

#include<bits/stdc++.h>
#include<sstream>

using namespace std;

#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'

const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);

    int cnt = 0;
    vector<string> res;
    string line;
    while(getline(cin, line)){
        stringstream ss(line);
        string word;
        bool ok = false;
        while(ss >> word){
            if(word == "<img"){
                cnt++;
                ok = true;
            }
            else if(ok == false) break;
            else if(word.size() >= 5 && word.substr(0, 4) == "src=")
                {res.push_back(word.substr(5, word.size() - 6));break;}
        }
        ss.clear();
    }
    cout << cnt << '\n';
    for(auto x : res) cout << x << '\n';
    return 0;
}

E.平方和

题面

思路

直接模拟即可

示例代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'

const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);

    ll ans = 0;
    fer(i, 1, 2020){
        int j = i;
        bool f = false;
        while(i){
            if(i % 10 == 2 || i % 10 == 0 || i % 10 == 1 || i % 10 == 9)
                {f = true; break;}
            i /= 10;
        }
        if(f) ans += j * j;
        i = j;
    }
    cout << ans << '\n';
    return 0;
}