2025“钉耙编程”中国大学生算法设计春季联赛（2）（2 4 5，补3 6 8)

2025“钉耙编程”中国大学生算法设计春季联赛（2）（2 4 5，补3 6 8)

1002：烂人方法

#include<bits/stdc++.h>
#define N 1005
#define mod 998244353
using namespace std;
typedef long long ll;
string a[] = {"jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui","jia","yi","bing","ding","wu","ji","geng","xin","ren","gui"};
//string b[] = {"zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai""zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai","zi","chou","yin","chen","si","wu","wei","shen","you","xv","hai"};
string b[] = {"zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai","zi","chou","yin","mao","chen","si","wu","wei","shen","you","xu","hai"};
void solve()
{string s;cin>>s;int i = 0;while(1){string c;c =  a[i]+b[i];//cout<<c;if(c == s){cout<<1984+i<<"\n";return;}i++;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);int T = 1;cin>>T;while(T--)solve();return 0;
}

1004:暴力dp

#include<bits/stdc++.h>
#define N 1005
#define mod 998244353
using namespace std;
typedef long long ll;
int f[10005];
void solve()
{string s1,s2;cin>>s1>>s2;int u = 0;if(s2.size() <= 2){while(!s2.empty()){u*=10;u += s2[0]-'0';s2.erase(0,1);}if(u >= 26){map<char,int>mp;for(int i = 0;i < s1.size();i++){if(mp[s1[i]])   mp[s1[i]]++;}cout<<mp.size()<<"\n";}else{string s;while(u--){s+=s1;}int ans = 0;for(int i = 0;i < s.size();i++){f[i] = 1;for(int j = 0;j < i;j++){if(s[j] < s[i]) {f[i] = max(f[i],f[j]+1);}}}for(int i = 1;i < s.size();i++) ans = max(ans,f[i]);cout<<ans<<"\n";}}else{map<char,int>mp;for(int i = 0;i < s1.size();i++){if(mp[s1[i]])   mp[s1[i]]++;}cout<<mp.size()<<"\n";}
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);int T = 1;cin>>T;while(T--)solve();return 0;
}

1005:理性分析

idea:

如果 ( k ) 是奇数，则答案为 0。

先讨论 ( k = 2 ) 的情形，此时有 ( \lceil n/2 \rceil ) 条平行 ( l_0 ) 的直线和 ( \lfloor n/2 \rfloor ) 条垂直 ( l_0 ) 的直线，答案就是二者的积。

对于 ( k > 2 ) 的情形，可以把所有 ( l_i ) 按照 ( i \mod (k/2) ) 分组，此时有 ( n \mod (k/2) ) 组直线有 ( \lfloor n/k \rfloor + 1 ) 条，另外 ( (k/2) - (n \mod (k/2)) ) 组直线有 ( \lfloor n/k \rfloor ) 条。显然垂直的直线都在同一个组内，并且每一个组自身都是一个 ( k = 2 ) 的情形，套用上面的结论即可。

#include<bits/stdc++.h>
#define N 1005
#define mod 998244353
using namespace std;
typedef long long ll;void solve()
{int n,k;cin>>n>>k;if(k%2!=0){cout<<0<<"\n";return ;}ll m = k/2;if(m > n-1){cout<<0<<"\n";return ;}ll q = (n-1)/m;ll maxj = (q-1)/2;if(maxj < 0){cout<<0<<"\n";return ;}ll cnt = maxj+1;ll ans = cnt*n-m*cnt*cnt;cout<<ans<<"\n";
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);int T = 1;cin>>T;while(T--)solve();return 0;
}

1003：

idea:关键在于找到第三个数字使得（p,0,p,q）能够为答案做出贡献，所以可以用从后往前map来记录一个数字首次出现的位置。再次从前往后遍历，如果p出现了超过两次，那么拿他首次出现位置和目前0的位置比较，如果x[p]<0，那么他就对答案做出贡献，其贡献即为当前>0的数字的后缀和，同时把当前的x[p]标记为inf，那么后面再次遍历到p的时候其则没有贡献

注意:这里的关键从后往前遍历找p首先出现的位置以及从前往后遍历0的位置，如果p只出现一次（后面即使有0,mp[a[i]]也不会再次出现)或者0的位置没有更新，那么p对答案不产生贡献

#include<bits/stdc++.h>
#define mod 998244353
using namespace std;
typedef long long ll;
const int N = 1e9+7;
int cnt[1000005];
void solve()
{ll n;memset(cnt,0,sizeof(cnt));cin>>n;vector<int>a(n+1);//1 1 1 1 0 1 3    1 0 1 2      2 0 2 2        3 5 7 8 9 1 0 2 2 0 0  for(int i = 1;i <= n;i++)    cin>>a[i];unordered_map<int,int>mp;int last0 = 0;vector<int>ans(n+2,0);for(int i = n;i >= 1;i--){if(a[i] == 0){ans[i] = ans[i+1];continue;}ans[i] = ans[i+1] + (mp[a[i]]==0);//cnt[a[i]] = i;mp[a[i]] = i;}ll res = 0;for(int i = 1;i <= n;i++){if(a[i] == 0){last0 = i;continue;}if(mp[a[i]] < last0){res += ans[i+1];mp[a[i]] = N;}}cout<<res<<"\n";
}
signed main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);int T = 1;cin>>T;while(T--)solve();return 0;
}

1006:博弈论，guess题

推荐佬的博客【杭电多校比赛记录】2025“钉耙编程”中国大学生算法设计春季联赛（2） - 天天超方的 - 博客园

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;void solve()
{ll n,m,k;cin>>n>>m>>k;if((n+2*m)%4)	cout<<"Alice\n";else	cout<<"Bob\n";
}
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr),cout.tie(nullptr);int T = 1;cin>>T;while(T--){solve();}
}

1008:拓扑，图

ieda:首先每个点只有3种可能的颜色

至于怎么推的我也guess了半天，如果某个块出现了直接no即可

如果都成立，那么这个点一定是属于后面的状态（即是最后一次经过的）此时只需要从另外两种颜色向这个颜色连一条边，表示这个点是后面经过的

最后进行拓扑排序，如果存在环，那么则不满足条件

#include<bits/stdc++.h>
#define N 100005
#define mod 998244353
using namespace std;
typedef long long ll;
#define int long longvoid solve()
{int n,k;cin>>n>>k;vector<vector<int>>g(3*n+1);vector<int>in(3*n+1,0);bool ok = 1;    while(k--){int x,y,z;cin>>x>>y>>z;int q = (y+1)/2,p = 2*n-x+1,r = 2*n+x-y/2;if(!ok) continue;if(z!=q && z!=p && z!=r){ok = 0;continue;}if(z == q){g[r].push_back(z);g[p].push_back(z);}else if(z == p){g[q].push_back(z);g[r].push_back(z);}else if(z == r){g[q].push_back(z);g[p].push_back(z);}in[z] += 2;}if(!ok){cout<<"No\n";return ;}auto tuopu = [&](){queue<int>q;for(int i = 1;i <= 3*n;i++){if(!in[i]){q.push(i);//找到入度为0的点}}while(!q.empty()){int now = q.front();q.pop();for(auto &i:g[now]){if((--in[i]) == 0){q.push(i);}}}for(int i = 1;i <= 3*n;i++){if(in[i])   return false;}return true;};if(tuopu()) cout<<"Yes\n";else    cout<<"No\n";
}
signed main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);int T = 1;cin>>T;while(T--)solve();return 0;
}