Codeforces Round #741 (Div. 2) ABCD1D2题解

Posted 2022-12-10 quinn18

文章目录

• A - The Miracle and the Sleeper
• B - Scenes From a Memory
• C - Rings
• D1 - Two Hundred Twenty One (easy version)
• D2 - Two Hundred Twenty One (hard version)
• 总结

---

A - The Miracle and the Sleeper

添加链接描述
题意：$l,r$ 里找两个数 $a>=b$ 使 $a$%$b$最大

好慢啊
想到 $(x+(x-1))$ % $x==x-1(x+x-1<=r)$
然后判断 $x$ 和 $l$ 哪个大 直接输出 $r$ % $max(x,l)$
%$l$确实是最大的 qwq

#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main() 
    int t;
    cin>>t;
    while(t--) //x+x+1
        int l, r;
        cin>>l>>r;
        int x=r/2+1;
        x=max(l, x);
        cout<<r%x<<endl;
    
	return 0;

B - Scenes From a Memory

添加链接描述
题意：给你个 $n$ 求删掉最多多少个数使这个数变成非质数或 $1$

好怂啊
找里面一些数字凑成非质数或者有 $1$ 就行
其实只要猜两位数就好了然后整了个素数筛ahhh

#include<bits/stdc++.h>
#define int long long
using namespace std;
bool vis[1000005];
int pri[1000005], cnt, n; 
void get_prime(int n) 
    vis[1] = 1;
    for (int i = 2; i <= n; i++) 
        if (!vis[i]) 
            pri[cnt++] = i;
            for (int j = 2 * i; j <= n; j += i) 
                vis[j] = 1;
            
        
    

string f(int m) 
    string w;
    while(m) 
        w=char('0'+m%10)+w;
        m/=10;
    
    return w;

int ct=0;
string ww[100005];
signed main() 
    get_prime(10005);
    for(int i=1; i<=10005; i++) 
        if(vis[i]) 
            ww[++ct]=f(i);
        
    
    int t;cin>>t;
    while(t--) 
        int m;cin>>m;
        string s;
        cin>>s;
        int q=0;
        for(int i=1; i<=ct; i++) 
            int sum=0;
            //cout<<ww[i]<<endl;
            for(int j=0; j<m; j++) 
                if(s[j]==ww[i][sum]) sum++;
                int len=ww[i].size();
                if(sum==len) 
                    cout<<len<<endl;
                    cout<<ww[i]<<endl;
                    q=1;
                    break;
                
             if(q)break;
        
    
	return 0;

C - Rings

添加链接描述
题意：给你一个二进制 截取两个比一半还长的二进制转换为十进制，截取区间不能一样，这两要是倍数关系

乌鱼子被吓唬了hhh
看了好久
先能发现 $<<1$ 是 $\ast 2$ 的意思是二进制左移一位，所以两个长一样的二进制一个后面加个 $0$ 就是倍数关系
然后想到 $k==1$ 的情况就 $0111$ 和 $111$

#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main() 
    int t;
    cin>>t;
    while(t--) 
        int n;
        cin>>n;
        string s;
        cin>>s;
        int sum=0;
        int q=0;
        for(int i=0; i<n; i++) if(s[i]=='1') sum++;
        if(!sum) //全0
            cout<<1<<" "<<n-1<<" "<<2<<" "<<n<<endl;
            continue;
        
        for(int i=0; i<n; i++) //找满足长度的0位置
            if(s[i]=='0') 
                if(i>=n/2) 
                    cout<<1<<" "<<i+1<<" "<<1<<" "<<i<<endl;
                    q++;
                    break;
                else
                    cout<<i+1<<" "<<n<<" "<<i+2<<" "<<n<<endl;
                    q++;
                    break;
                
            
        
        if(!q)//全1
            cout<<1<<" "<<n-1<<" "<<2<<" "<<n<<endl;
    
	return 0;

D1 - Two Hundred Twenty One (easy version)

添加链接描述
题意：给你个+ - 串 使$a1-a2+a3-a4+...=0$ $q$次问 最少移去多少个杆子使给你区间==0

就是摸你+前缀和

int sum[N];
signed main() 
    int t; cin>>t;
    while(t--) 
        int n, q; cin>>n>>q;
        string s; cin>>s;
        s='.'+s;
        int x=0, k=0;
        for(int i=1; i<=n; i++) 
            if(i%2==0) 
                if(s[i]=='+') sum[i]=sum[i-1]-1;
                else sum[i]=sum[i-1]+1;
            else 
                if(s[i]=='+') sum[i]=sum[i-1]+1;
                else sum[i]=sum[i-1]-1;
            
        
        for(int p=1; p<=q; p++) 
            int l, r; cin>>l>>r;
            if((r-l+1)%2==0) 
                if((sum[r]-sum[l-1])==0) cout<<0<<endl;
                else cout<<2<<endl;
            else 
                cout<<1<<endl;
            
        
    
	return 0;

D2 - Two Hundred Twenty One (hard version)

添加链接描述

每次A完一题就谢了qwq就没了

---

总结