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 << '\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 << '\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 << '\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 << '\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 << '\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;
}
