概要
L1一共8题,5分,10分,15分,20分各两题。
5分题一般会输入输出就行,10分题就是一个循环或者选择结构
15分题有简单的模拟和查找,20分题是稍微繁琐的简单模拟这样。
L1-073 人与神 5 104 135 0.77
L1-074 两小时学完C语言 5 95 124 0.77
L1-075 强迫症 10 68 137 0.50
L1-076 降价提醒机器人 10 66 89 0.74
L1-077 大笨钟的心情 15 58 126 0.46
L1-078 吉老师的回归 15 68 190 0.36
L1-079 天梯赛的善良 20 69 188 0.37
L1-080 乘法口诀数列 20 71 165 0.43
L1-073 人与神 (5分)
#include<bits/stdc++.h>using namespace std;int main(){cout<<"To iterate is human, to recurse divine.\n";return 0;}
L1-074 两小时学完C语言 (5分)
#include<bits/stdc++.h>using namespace std;int main(){int n, k, m;cin>>n>>k>>m;int x = n-k*m;if(x<0)x = 0;cout<<x<<"\n";return 0;}
L1-075 强迫症 (10分)
#include<bits/stdc++.h>using namespace std;int main(){string s;cin>>s;if(s.size()==4){int x = stoi(s.substr(0,2));if(x<22){cout<<20<<s.substr(0,2)<<"-"<<s.substr(2,2)<<"\n";}else{cout<<19<<s.substr(0,2)<<"-"<<s.substr(2,2)<<"\n";}}else{cout<<s.substr(0,4)<<"-"<<s.substr(4,2);}return 0;}
L1-076 降价提醒机器人 (10分)
#include<bits/stdc++.h>using namespace std;int main(){int n; double m; cin>>n>>m;for(int i = 1; i <= n; i++){double p; cin>>p;if(p<m)printf("On Sale! %.1lf\n",p);}return 0;}
L1-077 大笨钟的心情(15分)
#include<bits/stdc++.h>using namespace std;int a[50];int main(){for(int i = 0; i < 24; i++)cin>>a[i];int x; while(cin>>x&&x>=0&&x<=23){if(a[x]>50)cout<<a[x]<<" Yes\n";else cout<<a[x]<<" No\n";}return 0;}
L1-078 吉老师的回归(15分)
#include<bits/stdc++.h>using namespace std;int a[50];int main(){int n, m;cin>>n>>m; cin.get();int cnt = 0;for(int i = 1; i <= n; i++){string s; getline(cin,s);if(s.find("easy")==string::npos && s.find("qiandao")==string::npos){//cout<<"adsfkragnnlkm";cnt++;if(cnt==m+1){cout<<s<<"\n";return 0;}}}cout<<"Wo AK le\n";return 0;}
L1-079 天梯赛的善良 (20分)
#include<bits/stdc++.h>using namespace std;const int maxn = 2e4+10;int a[maxn];int main(){int n; cin>>n;int mx = -1, mi = 1e9+10;for(int i = 1; i <= n; i++){cin>>a[i]; mx = max(mx, a[i]);mi = min(mi, a[i]);}int a1 = 0, a2 = 0;for(int i = 1; i <= n; i++){if(a[i]==mx)a1++;if(a[i]==mi)a2++;}cout<<mi<<" "<<a2<<"\n";cout<<mx<<" "<<a1<<"\n";return 0;}
L1-080 乘法口诀数列 (20分)
#include<bits/stdc++.h>using namespace std;const int maxn = 2e4+10;int a[maxn];int main(){int n;cin>>a[1]>>a[2]>>n;int cur = 3;for(int i = 3; i <= n; i++){int x = a[i-1]*a[i-2];string s = to_string(x);for(int j = 0; j < s.size(); j++){a[cur++] = s[j]-'0';}}for(int i = 1; i <= n; i++){if(i!=n)cout<<a[i]<<" ";else cout<<a[i];}return 0;}