2021-11-27【NEC Programming Contest 2021(AtCoder Beginner Contest 229)】【A-D题解】

#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair pii;

const int maxn=1e6+5;

int main(){
    ios::sync_with_stdio(0);
    char a[4][4] = {0};
    int n = 2;
    for (int i = 0; i < n;i++){
        for (int j = 0; j < n;j++){
            cin >> a[i][j];
        }
    }
    int vis[3][3] = {0};
    for (int i = 0; i < n;i++){
        for (int j = 0; j < n;j++){
            if(a[i][j]=='#')
                vis[i][j] = 1;
        }
    }
    int flag = 0;
    for (int i = 0; i < n;i++){
        int sum = 0;
        for (int j = 0; j < n;j++){
            sum += vis[i][j];
        }
        //cout << sum << endl;
        if(sum>1){
            flag = 1;
        }
    }
    for (int i = 0; i < n;i++){
        int sum = 0;
        for (int j = 0; j < n;j++){
            sum += vis[j][i];
        }
        //cout << sum << endl;
        if(sum>1){
            flag = 1;
        }
    }
    if(!flag)
        cout << "No" << endl;
    else
        cout << "Yes" << endl;
    return 0;
}

#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair pii;

const int maxn=1e6+5;
void solve(){
    string a, b;
    int flag = 0;
    cin>>a>>b;
    for (int i = 0; i < min(a.size() , b.size());i++){
        if(a[a.size()-1-i]-'0'+b[b.size()-1-i]-'0'>=10)
            flag = 1;
    }
    if(flag)
        cout << "Hard";
    else
        cout << "Easy";
}

int main(){
    ios::sync_with_stdio(0);
        solve();
    return 0;
}

#include 
#define ll long long
using namespace std;
ll n, w, ans;
const int maxn=1e6+10;
void solve(){
	cin>>n>>w;
	pair< ll,ll > a[maxn];
	for(int i=1;i<=n;i++){
		cin>>a[i].first>>a[i].second;
	}
	sort(a+1,a+n+1);
	for(int i=n;i>=1;i--){
		ans+=a[i].first*min(a[i].second,w);
		w-=min(w,a[i].second);
	} 
	cout< 
D - Longest X 
 
题意：给出一个字符串只含有X和.给定k步，每一步将.变为X，求k步后最长连续X的长度。 
解析：每次先向右找k个.，然后记录最大X长度，之后移动左端点，再继续移动右端到满足k个.直到字符串的结尾。 
#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair pii;

const int maxn=1e6+5;
void solve(){
    string s;
    cin>>s;
    int n = s.size();
    int k;
    cin >> k;
    int r = 0, cur = 0, ans = 0;
    for (int i = 0; i < n;i++){
        while(r 
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