A. Antiamuny wants to learn binary search

签到题.

#include  <map>
#include  <set>
#include  <cmath>
#include  <queue>
#include  <stack>
#include  <cstdio>
#include  <vector>
#include  <climits>
#include  <cstring>
#include  <cstdlib>
#include  <iostream>
#include  <algorithm>
#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 1e9 + 10, mod = 1e9 +7;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<char, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*

 */
string s;
vector<PII> a,ans;
int f(int l,int r,int x) { // l <= x <= r
    int cnt = 0;
    while(l <= r) {
        cnt++;
        int mid = (l + r) / 2;
        if (mid == x) break;
        if (mid < x) l = mid + 1;
        else r = mid - 1;
    }
    return cnt;
}
void solve()
{
    cin >> n >> m >> k;
    cout << f(n,m,k) << endl;

}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    cin >> Ke_scholar ;
    while(Ke_scholar--)
        solve();
    return 0;
}

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B. Bespread with chequers

用dp数组来代表n列时的方案总数,可以发现n是偶数时,dp[n] = dp[n - 1] + 1, 奇数时dp[n] = dp[n - 1] - 1,别忘了取模.

#include<bits/stdc++.h>

#define int long long
#define endl '\n'
#define fi first
#define se second

using namespace std;

const int N = 1e6 + 10,mod = 1e9 + 7;

int n, m, x;
int dp[N];
void init(){
    dp[1] = 1;
    for(int i = 2;i < 1e6 + 10;i ++){
        if(i & 1)
            dp[i] = (2 * dp[i - 1] % mod - 1) % mod;
        else
            dp[i] = (2 * dp[i - 1] % mod + 1) % mod;
    }
}

void solve() {
    cin >> n;
    cout << dp[n] % mod << endl;
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int h_h;
    init();
    cin >> h_h;
    //h_h = 1;
   for(int i=1;i<=h_h;i++)solve();
    return 0;
}

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C.Converse the string

即就是对字符串s遍历一遍,用s[i]和字符串ans的首字符比较,字典序小的放前面,反之放后面.

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 1e6+10, mod = 1e9 + 7;

//typedef long long ll;
typedef pair<int,int> PII;
int n,m,t,k;
map<int,int> mp;
priority_queue<int> QQ;
deque<int> Q;
void solve() {
    string s;
    cin >> s;
    deque<char> ans,ori;
    for(auto i : s)
        if(ans.empty())
            ans.push_back(i);
        else{
            if(i <= ans.front()){
                ans.push_front(i);
        }
            else{
                ans.push_back(i);
            }
    }
    for(auto i : ans){
        cout << i ;
    }
    cout << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}
/*

 */

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G.Golden jade matrix checker

二维前缀和,对给出x,y矩阵,去原矩阵中找他的和即可,大于等于0的直接输出No即可.

#include<bits/stdc++.h>

#define int long long
#define endl '\n'
#define fi first
#define se second

using namespace std;

const int N = 2010;
const int M = 20;

int a[N][N];

void solve() {
    int n, m, x, y;
    cin >> n >> m >> x >> y;
    vector<vector<int>> s(n + 1,vector<int> (m + 1,0));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j];
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
        }
    }
    if (s[n][m] <= 0) {
        cout << "NO" << endl;
        return;
    }
    for(int i = 1;i <= n - x + 1;i ++){
        for(int j = 1;j <= m - y + 1;j++){
            int dx = i + x, dy = j + y;
            int res = s[dx - 1][dy - 1] - s[dx-1][j - 1] - s[i - 1][dy-1] + s[i - 1][j - 1];
            if(res >= 0){
                cout << "NO" << endl;
                return ;
            }
        }
    }
    cout << "YES" << endl;
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int h_h;
    cin >> h_h;
    //h_h = 1;
    while (h_h--)solve();
    return 0;
}

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H.How to know the function

很容易能找出,只有n为0时一次即可,其余次数均可以2次求出,记得开longlong.

#include<bits/stdc++.h>
#define endl '\n'
using namespace std;

const int N = 501;
long long n,m,x;
char mp[N][N];
int main(){
    int T;
    cin>>T;
    while(T--){
        cin >> n;
        if( n == 0){
            cout << 1 << endl;
        }
        else {
            cout << 2 << endl;
        }
    }
    return 0;
}

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J. Juxtaposed brackets

由定义可知,每个()里最多包含两个并列的括号,被括号包含的括号算是一个,例如(()()),外面的括号就是包含了两个并列的,而((()))这样只算包含了一个,如此,我们可以开一个数组去记录每一个括号内包含了多少个并列的括号,其中不对称,或者左扩右扩不相等的都可以做个标记,之后输出NO即可,对付符合要求的,再对数组进行遍历,若有大于2的则说明有括号包含了2个以上的并列括号,输出NO.

#include  <map>
#include  <set>
#include  <cmath>
#include  <queue>
#include  <stack>
#include  <cstdio>
#include  <vector>
#include  <climits>
#include  <cstring>
#include  <cstdlib>
#include  <iostream>
#include  <algorithm>
#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 1e9 + 10, mod = 1e9 +7;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<char, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*

 */
string s;
vector<PII> a,ans;
void solve()
{
    string s;
    cin >> s;
    bool f = false;
    vector<int> a(s.size());
    stack<int> st;
    for(int i = 0;i < s.size();i ++){
        if(s[i] == '('){
            st.push(i);
        }
        else if(st.empty()){
            f = true;
            break;
        }
        else {
            st.pop();
            if(st.size())
                a[st.top()]++;
        }
        if(st.empty() && i < s.size() - 1){
            f = true;
            break;
        }
    }
    for(auto i : a){
        if(i > 2){
            f = true;
            break;
        }
    }
    if(f || st.size()){
        cout << "NO" << endl;
    }
    else
        cout << "YES" << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    cin >> Ke_scholar ;
    while(Ke_scholar--)
        solve();
    return 0;
}