记录历年CCF CSP中的中档题（第三题）

201503-3

节日

#include <iostream>
using namespace std;
const int N = 1e6 + 10;

int a, b, c, y1, y2;
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int rec[3000][15][40];

bool is_leap(int x)
{
    if ((x % 4 == 0 && x % 100) || x % 400 == 0) return true;
    return false;
}

int main()
{
    cin >> a >> b >> c >> y1 >> y2;
    
    int ini = 2;
    
    for (int i = 1850; i <= 2050; i ++ )
    {
        for (int j = 1; j <= 12; j ++ )
        {
            int days = month[j];
            if (is_leap(i) && j == 2) days ++;
            for (int k = 1; k <= days; k ++ )
            {
                rec[i][j][k] = ini ++ % 7;
            }
        }
    }
    
    for (int i = y1; i <= y2; i ++ )
    {
        bool flag = false;
        if (c == 7) c = 0;
        int days = month[a];
        if (is_leap(i) && a == 2) days ++;
        int cnt = 0;
        for (int k = 1; k <= days; k ++ )
        {
           
            if (rec[i][a][k] == c) cnt ++;
            if (cnt == b)
            {
                printf("%d/%02d/%02d\n", i, a, k);
                flag = true;
                break;
            }
        }
        if (!flag) puts("none");
    }
    
    return 0;
}

 

 

 

 

 

201412-3

集合竞价

#include <iostream>

using namespace std;
const int N = 1e6 + 10;

int n, mx, tmp, b[N];
double a[N];
string op[N];
long long res, by[N], sl[N];

int main()
{
    for (int i = 1; ; i ++ )
    {
        if (cin >> op[i]) n ++;
        else break;

        if (op[i][0] != 'c') cin >> a[i] >> b[i];
        else 
        {
            cin >> tmp;
            op[tmp] = 'd';
        }
    }

    for (int i = 1; i <= n; i ++ )
    {
        tmp = a[i] * 1000 / 10;
        if (op[i][0] == 'b') by[tmp] += b[i];
        if (op[i][0] == 's') sl[tmp] += b[i];
    }

    for (int i = 1; i <= N; i ++ ) by[i] += by[i - 1], sl[i] += sl[i - 1];

    for (int i = 1; i <= N; i ++ ) 
        if (min(sl[i], by[N] - by[i - 1]) >= min(sl[mx], by[N] - by[mx - 1])) mx = i;

    printf("%.2lf %lld\n", mx / 100.0, min(sl[mx], by[N] - by[mx - 1]));
    return 0;
}

 

 

201409-3

字符串匹配

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 1010;

int n, st;
string S, str[N];
    
void lw(string& s)
{
    for (int i = 0; i < s.size(); i ++ )
        if (s[i] >= 'A' && s[i] <= 'Z') 
            s[i] = s[i] - 'A' + 'a';
}

bool find(string s)
{
    if (!st) lw(s);
    if (!st) lw(S);
    
    for (int i = 0; i < s.size(); i ++ )
        if (S == s.substr(i, S.size())) return true;
    return false;
}

int main()
{
    cin >> S >> st >> n;
    
    for (int i = 1; i <= n; i ++ ) 
        cin >> str[i];
    
    for (int i = 1; i <= n; i ++ )
        if (find(str[i])) cout << str[i] << endl;
    
    return 0;
}

 

 201403-3

命令行选项

#include <bits/stdc++.h>

using namespace std;

unordered_map<char, int> h;
unordered_map<char, int> uh;

int ck(string s)
{
    return s[0] == '-' && s.size() == 2;
}

bool find(string s, vector<string> &res)
{
    for (auto& x : res)
        if (x.substr(0, 2) == s.substr(0, 2)) return true;
    return false;
}

void rpc(string s, vector<string> &res)
{
    for (auto& x : res)
        if (x.substr(0, 2) == s.substr(0, 2)) x = s;
}

bool cmp(string a, string b)
{
    return a[1] < b[1];
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s;
    getline(cin, s);

    for (int i = 0; i < s.size(); i ++)
    {
        if (i + 1 < s.size() && s[i + 1] == ':')
        {
            h[s[i]] = 1;
            i ++;
        }
        else uh[s[i]] = 1;
    }

    getline(cin, s);
    int n = atoi(s.c_str());

    for (int i = 1; i <= n; i ++)
    {
        vector<string> now, res;
        int st = 0;
        string t;
        getline(cin, s);
        for (int j = 0; j < s.size(); j ++)
        {
            if (s[j] == ' ') 
            {
                st ++;
                if (t.size()) now.push_back(t);
                t.clear();
            }
            else if (st) t += s[j];
        }
        if (t.size()) now.push_back(t);
        t.clear();
        
        for (int j = 0; j < now.size(); j ++)
        {
            if (ck(now[j]))
                if (h[now[j][1]] && j + 1 < now.size() )
                {
                    if (find(now[j], res)) rpc(now[j] + ' ' + now[j + 1], res);
                    else res.push_back(now[j] + ' ' + now [j + 1]);
                    j ++;
                }
                else if (uh[now[j][1]] && !find(now[j], res)) 
                    res.push_back(now[j]);
                else break;
            else break;
        }
        sort(res.begin(), res.end(), cmp);
        cout << "Case " << i << ": ";
        for (auto x : res) cout << x << ' ';
            cout << endl;
    }
    return 0;
}

 

 

 

 

201312-3

最大的矩形

#include <iostream>
using namespace std;
const int N = 1010;

int n, res;
int q[N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ ) cin >> q[i];
    
    for (int i = 1; i <= n; i ++ )
    {
        int cnt = 1, j = i;
        while (j - 1 >= 1 && q[j - 1] >= q[i]) j --;
        cnt += i - j;
        j = i;
        while (j + 1 <= n && q[j + 1] >= q[i]) j ++;
        cnt += j - i;
        res = max(res, q[i] * cnt);
    }
    cout << res;
    return 0;
}