训练情况
赛后反思
这场怎么都是猜猜乐+典题，做不出来的题真的不会了
A题
显然 2025 里面有多少个 15,除法向上取整即可
点击查看代码
#include <bits/stdc++.h>
// #define int long long
#define endl '\n'using namespace std;void solve(){cout<<ceil(1.0*2025/15);
}signed main(){// int T; cin>>T; while(T--)solve();return 0;
}
B题
猜猜乐题，我们考虑贪心，如果当前位置已经在下一个区间就不移动，否则就移动到左右端点最近的那个，正确性我不会证明，手玩可知吧。。
点击查看代码
#include <bits/stdc++.h>
#define int long long
#define endl '\n'using namespace std;void solve(){int ans = 0;int n,m; cin>>n>>m;vector<int> x(n + 1),y(n + 1);int now = 1;for(int i = 1;i<=n;i++) cin>>x[i]>>y[i];for(int i = 1;i<=n;i++){if(now >= x[i] && now <= y[i]) continue;if(abs(now-x[i]) < abs(now-y[i])){ans += abs(now-x[i]);now = x[i];} else {ans += abs(now-y[i]);now = y[i];}}cout<<ans<<endl;
}signed main(){// int T; cin>>T; while(T--)solve();return 0;
}
C题
显然对于比一半数还小的，我们想改变答案，必须将大于一半的数加到小的数里面，但是我们发现一定会多一个 0，无论如何操作都不会让答案增加，所以我们直接对小于一半的数求和就是答案了
点击查看代码
#include <bits/stdc++.h>
#define int long long
#define endl '\n'using namespace std;void solve(){int n,k; cin>>n>>k;vector<int> a(n + 1);for(int i = 1;i<=n;i++) cin>>a[i];sort(a.begin() + 1,a.end());int ans = 0;for(int i = 1;i<=n/2;i++) ans += a[i];cout<<ans<<endl;
}signed main(){// int T; cin>>T; while(T--)solve();return 0;
}
D题
我们考虑暴力打表，枚举每一个位置放或不放，判断情况是否合法，合法则答案增加，我们发现有一个数列递推公式 a[i] = a[i-1] + (i-1)*a[i-2];
点击查看代码
#include <bits/stdc++.h>
// #define int long long
#define endl '\n'using namespace std;const int N = 100;int n;
int v[N*N];
int ans = 0;bool pd(){int c = 0;for(int i = 1;i<=n*n;i++) c += v[i];if(c!=n) return false;int a[N][N];for(int i = 1;i<=n;i++){for(int j = 1;j<=n;j++){a[i][j] = v[n*(i-1)+j];}}for(int i = 1;i<=n;i++){int cnt = 0;for(int j = 1;j<=n;j++){cnt+=a[i][j];}if(cnt>1) return false;}for(int i = 1;i<=n;i++){int cnt = 0;for(int j = 1;j<=n;j++){cnt+=a[j][i];}if(cnt>1) return false;}for(int i = 1;i<=n;i++){for(int j = 1;j<=i;j++){if(a[i][j] != a[j][i]) return false;}}// for(int i = 1;i<=n*n;i++) cout<<v[i]<<" "; cout<<endl;// for(int i = 1;i<=n;i++){// for(int j = 1;j<=n;j++){// cout<<a[i][j]<<" ";// }// cout<<endl;// }return true;
}int tot = 0;void dfs(int x){if(x > n*n){if(pd()) ans++;return;}for(int i = 0;i<=1;i++){if(tot==n&&i==1) continue;if(i == 1) tot++;v[x] = i;dfs(x + 1);if(i == 1) tot--;v[x] = 0;}
}void solve(){cin>>n;dfs(1);cout<<ans<<endl;
}signed main(){// int T; cin>>T; while(T--)solve();return 0;
}
#include <bits/stdc++.h>
#define int long long
#define endl '\n'using namespace std;const int mod = 1e9 + 7;
const int N = 1e6 + 3;int a[N];void pre(){a[0] = 1;a[1] = 1;for(int i = 2;i<=N-3;i++){a[i] = a[i-1] + (i-1)*a[i-2];a[i] %= mod;}
}void solve(){int x; cin>>x;cout<<a[x]<<endl;
}signed main(){pre();int T; cin>>T; while(T--)solve();return 0;
}
