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 = cntn-mcnt*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
for(int i = 1;i <= n;i++) cin>>a[i];
unordered_map<int,int>mp;
int last0 = 0;
vector
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
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 long
void solve()
{
int n,k;
cin>>n>>k;
vector<vector
vector
bool ok = 1;
while(k--)
{
int x,y,z;
cin>>x>>y>>z;
int q = (y+1)/2,p = 2n-x+1,r = 2n+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
for(int i = 1;i <= 3n;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 <= 3n;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;
}
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