2024 CSP-J第二轮复赛真题及参考答案

P11227 [CSP-J 2024] 扑克牌（民间数据）
#include <bits/stdc++.h>using namespace std;typedef long long ll;int n;set<string> st;string s;int main(){    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);    ll i, j;    cin>>n;    for(i=1; i<=n; i++)    {        cin>>s;        st.insert(s);    }    cout<<52-st.size();    return 0;}
P11228 [CSP-J 2024] 地图探险（民间数据）
#include <bits/stdc++.h>using namespace std;typedef long long ll;int t;int n, m, k;int x, y, d;string s;int a[1005][1005];
int main(){    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);    ll i, j;    cin>>t;    while(t--)    {        cin>>n>>m>>k;        cin>>x>>y>>d;        memset(a, 0, sizeof a);        for(i=1; i<=n; i++)        {            cin>>s;            s = ' ' + s;            for(j=1; j<=m; j++)            {                if(s[j] == '.')                    a[i][j] = 0;                else                    a[i][j] = -1;            }        }        a[x][y] = 1;        for(i=1; i<=k; i++)        {            int x1, y1;            if(d == 0)                x1 = x, y1 = y+1;            else if(d == 1)                x1 = x+1, y1 = y;            else if(d == 2)                x1 = x, y1 = y-1;            else if(d == 3)                x1 = x-1, y1 = y;            if(1<=x1 && x1 <= n && 1<=y1 && y1 <= m && a[x1][y1] >= 0)                x = x1, y = y1, a[x][y] = 1;            else                d = (d+1)%4;        }        int ans = 0;        for(i=1; i<=n; i++)            for(j=1; j<=m; j++)                if(a[i][j] == 1)                    ans++;        cout<<ans<<endl;    }    return 0;}
P11229 [CSP-J 2024] 小木棍（民间数据）
#include <bits/stdc++.h>using namespace std;typedef long long ll;int c[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};int ans[100100];int t, n;
int main(){    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);    ll i, j;    ans[1] = -1;    ans[2] = 1;    ans[3] = 7;    ans[4] = 4;    ans[5] = 2;    ans[6] = 6;    ans[7] = 8;    ans[8] = 10;    ans[9] = 18;    ans[10] = 22;    ans[11] = 20;    ans[12] = 28;    ans[13] = 68;    ans[14] = 88;    ans[15] = 108;    ans[16] = 188;    ans[17] = 200;    ans[18] = 208;    ans[19] = 288;    ans[20] = 688;    ans[21] = 888;    cin>>t;    while(t--)    {        cin>>n;        if(n<=20)            cout<<ans[n]<<endl;        else        {            int s = n/7-2;            int z = n%7+14;            cout<<ans[z];            for(i=1; i<=s; i++)                cout<<8;            cout<<endl;        }    }    return 0;}  P11230 [CSP-J 2024] 接龙（民间数据）
#include <bits/stdc++.h>using namespace std;typedef long long ll;int t;int n, k, q;int li[100100];vector<int>s[100100], ok[100100];struct dd{    int r, c;}dat[100100];bool vis[102][200100];int p[200100];
int main(){    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);    ll i, j;    int tp;    cin>>t;    while(t--)    {        cin>>n>>k>>q;        memset(vis, 0, sizeof vis);        for(i=1; i<=n; i++)        {            cin>>li[i];            s[i].clear();            s[i].push_back(0);            ok[i].clear();            ok[i].resize(li[i]+1);            for(j=1; j<=li[i]; j++)            {                cin>>tp;                s[i].push_back(tp);            }        }        for(i=1; i<=q; i++)        {            cin>>dat[i].r>>dat[i].c;        }        for(i=1; i<=n; i++)            for(j=1; j<=li[i]; j++)                if(s[i][j] == 1)                    ok[i][j] = 1;                else                    ok[i][j] = 0;        for(i=1; i<=100; i++)        {            for(j=1; j<=200000; j++)                p[j] = 0;            for(j=1; j<=n; j++)            {                int lst = -1e6;                for(int z=1; z<=li[j]; z++)                {                    if(z-lst+1 <= k)                    {                        if(p[s[j][z]] == 0)                            p[s[j][z]] = j;                        else if(p[s[j][z]] != j)                            p[s[j][z]] = 1e6;                        vis[i][s[j][z]] = 1;                        }                    if(ok[j][z]) lst = z;                }            }            for(j=1; j<=n; j++)            {                for(int z=1; z<=li[j]; z++)                {                    if(p[s[j][z]] && p[s[j][z]] != j)                        ok[j][z] = 1;                    else                        ok[j][z] = 0;                }            }        }        for(i=1; i<=q; i++)        {            if(vis[dat[i].r][dat[i].c])                cout<<1<<endl;            else                cout<<0<<endl;        }    }    return 0;}