Codeforces Round #806 (Div. 4)

A. YES or YES?

思路：algorithm库里有一个transform函数可以将字符串转为大写，然后判断是否等于YES

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int a[N];

void solve()
{
    cin >> n;
    while (n--)
    {
        string s;
        cin >> s;
        transform(s.begin(), s.end(), s.begin(), ::toupper);
        if (s != "YES")
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
}

int main()
{
    int T = 1;
    //cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}

B. ICPC Balloons

思路：当一个字母首次出现的时候加2，不是首次出现就加1

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int a[N];

void solve()
{
    cin >> n;
    set<char> s;
    int res = 0;
    for (int i = 1; i <= n; i ++ )
    {
        char ch;
        cin >> ch;
        if(s.count(ch))
            res += 1;
        else
            res += 2;
        s.insert(ch);
    }
    cout << res << endl;
}

int main()
{
    int T = 1;
    cin >> T;
    while(T -- )
    {
        solve();
    }
    return 0;
}

C. Cypher

思路：反向操作，给出的字符串U就往下转，即减1，D就往上转，即加1注意0往下转和9往上转的情况

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int a[N];

void solve()
{
    cin >> n;
    int res[110];
    for (int i = 1; i <= n; i ++ )
    {
        cin >> res[i];
    }
    for (int i = 1; i <= n; i ++ )
    {
        int cnt;
        cin >> cnt;
        while(cnt -- )
        {
            char op;
            cin >> op;
            if(op == 'U')
            {
                if(res[i] == 0)
                    res[i] = 9;
                else res[i]--;
            }
            else 
            {
                if(res[i] == 9)
                    res[i] = 0;
                else
                    res[i]++;
            }
        }
        cout << res[i] << ' ';
    }
    cout << endl;
}

int main()
{
    int T = 1;
    cin >> T;
    while(T -- )
    {
        solve();
    }
    return 0;
}

D. Double Strings

思路：暴力，因为每个字符串最多就8位，从第一个字母开始从前向后分成两部分，找前后两部分有没有

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 100010;

int n, m;
int a[N];

void solve()
{
    cin >> n;
    set<string> s;
    vector<string> v;
    string res = "";
    for (int c = 1; c <= n; c ++ )
    {
        string str;
        cin >> str;
        s.insert(str);
        v.push_back(str);
    }
    for (int i = 0; i < v.size(); i ++ )
    {
        bool flag = false;
        string tmp = "";
        for (int j = 0; j < v[i].size(); j ++ )
        {
            tmp += v[i][j];
            if(tmp == v[i])
                continue;
            if(s.count(tmp))
            {
                string tmp2 = "";
                for (int k = j + 1; k < v[i].size(); k ++ )
                    tmp2 += v[i][k];
                if(s.count(tmp2))
                {
                    res += "1";
                    flag = true;
                    break;
                }
            }
        }
        if(!flag)
            res += "0";
    }
    cout << res << endl;
}

int main()
{
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}

F. Yet Another Problem About Pairs Satisfying an Inequality

思路：分成三个不等式来看，ai<i, aj<j, i<aj,  然后可以先把ai<i的数存起来，然后就只需要找满足i<aj的数就可以了，这里可以用lower_bound，找某一个存起来的数前面有多少个数的坐标比它的值小

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int N = 200010;

int n, m;
struct node{
    LL val, idx;
    bool less;
    bool operator< (const node& t) const{
        return idx < t.idx;
    }
} a[N];
int sum[N];

void solve()
{
    cin >> n;
    set<node> s;
    vector<LL> v;
    for (int i = 1; i <= n; i ++ )
    {
        cin >> a[i].val;
        a[i].idx = i;
        if(a[i].val - a[i].idx >= 0)
            a[i].less = false;
        else
            a[i].less = true;

        if(a[i].less)
        {
            s.insert(a[i]);
            v.push_back(a[i].idx);
        }
    }
    sort(v.begin(), v.end());

    // for (int i = 0; i < v.size(); i ++ )
    //     cout << v[i].val << ' ' << v[i].idx << endl;

    LL res = 0;
    for(auto x: s)
    {
        //v[j].val -> v[i].idx
        LL tmp = lower_bound(v.begin(), v.end(), x.val) - v.begin();
        res += tmp;
    }


    cout << res << endl;
}

int main()
{
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}

标签：typedef,int,res,Codeforces,long,--,solve,Div,806
来源： https://www.cnblogs.com/yctql/p/16475734.html