冲刺DDDDDAAAAAYYYYY11111

Posted on By 孙 绍睿 冲刺DDDDDAAAAAYYYYY11111无评论




第一题

#include<bits/stdc++.h>
using namespace std;
int n;
char mp[114][114];
int main(){
    freopen("titato.in","r",stdin);freopen("titato.out","w",stdout);
    cin>>n;
    if(n>26){
        cout<<"N0";
        return 0;
    }
    for(int i=0;i<n;i++){
        cin>>mp[i];
    }
    for(int i=0;i<n;i++){
        bool ltr[26]={0},flag=false;//whether have the same letter
        for(int j=0;j<n;j++){
            if(ltr[mp[i][j]-&#039;a&#039;]==true){
                flag=true;
                break;
            }
            ltr[mp[i][j]-&#039;a&#039;]=true;
        }
        if(flag==false){
            cout<<"YE5";
            return 0;
        }
    }
    for(int i=0;i<n;i++){
        bool ltr[26]={0},flag=false;//whether have the same letter
        for(int j=0;j<n;j++){
            if(ltr[mp[j][i]-&#039;a&#039;]==true){
                flag=true;
                break;
            }
            ltr[mp[j][i]-&#039;a&#039;]=true;
        }
        if(flag==false){
            cout<<"YE5";
            return 0;
        }
    }
    cout<<"N0";
    return 0;
}
很容易想到：
如果方阵边长大于26，就一定会有重复的，直接输出“N0”即可
有两个坑点：
输出的是“N0”“YE5”(嗯零，歪易五)而不是“NO”“YES”（嗯哦，歪易诶斯）
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第二题
#include<bits/stdc++.h>
using namespace std;
long long qst;
int main(){
    freopen("bricks.in","r",stdin);freopen("bricks.out","w",stdout);
    cin>>qst;
    while(qst--){
        long long a,b,c,d;
        cin>>a>>b>>c>>d;
        if(b==d){
            if(a==c)cout<<"T 1\n";
            else cout<<"F 1\n";
            continue;
        }else{
            b++;
            if(c>=a-d+b&&c<=a+d-b)cout<<"T ";
            else cout<<"F ";
            cout<<(d-b+1)*2<<&#039;\n&#039;;
        }
    }
    return 0;
}
（改了好久才改出来的）
在考试的时候就想到可以把图画出来，找规律然后O(1)求解
其实就是if来if去的
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第三题
做法明白力，但是代码打不出来，所以代码就不放出来力
先把完全二叉树用上面的公式转成满二叉树
把叶子节点全部加起来
一直除以二除以二除上去就可以得到根节点的横坐标力
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别看了，后面两题我都不会


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