天梯赛6and模拟赛

7-87 检查密码 (15分)

题意:就是根据给的字符串判断,进行输出。

题解:就是根据字符串的遍历找到是否有数字或者字母进行标记,特别主要要特判都是点的情况,需要输出完美。

代码:

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    string s;
    getchar();
    int tag1=0,tag2=0,tag3=0,tag4=0;
    while(n--){
        getline(cin,s);
          if(s.length()<6){
            cout<<"Your password is tai duan le."<<endl;
        }else {
        for(int i=0;i<s.length();i++){
            if(s[i]>='0'&&s[i]<='9'){
                tag1=1;
            }
            else if((s[i]>='a'&&s[i]<='z')||(s[i]>='A'&&s[i]<='Z')){
                tag2=1;
            }
            else if(s[i]=='.'){
                tag3=1;
            }
            else{
                tag4=1;
            }
        }
      if(tag4==1){
            cout<<"Your password is tai luan le."<<endl;
        }else if(tag2==1&&tag1==0){
            cout<<"Your password needs shu zi."<<endl;
        }else if(tag1==1&&tag2==0){
            cout<<"Your password needs zi mu."<<endl;
        }
        else{
            cout<<"Your password is wan mei."<<endl;
        }
        }
        tag1=0,tag2=0,tag3=0,tag4=0;
    }
    return 0;
}

7-9 关于堆的判断 (25分)

题意:给定序列和输入判断是对的还是错的。

题解:主要难点是小顶堆的建立,只要把小顶堆建立好了,然后还要知道树的定义a[k]的左子树是a[k/2]右子树是a[k/2+1],再根据输入的字符串判断条件即可。

代码:

#include <iostream>
using namespace std;
const int MAX = 1002;
int a[MAX];
int n,m;
int find(int x){
    for (int i = 1; i <= n;i++){
        if(a[i]==x) return i;
    }
    return 0;
}
int main(){
    cin >> n >> m;
    for (int i = 1; i <= n;i++){
        cin >> a[i];
        int k=i;
        while(k>1&&a[k]<a[k/2]){//小顶堆的概念,建立小顶堆 
            swap(a[k],a[k/2]);
            k/=2;
        }
    }
//    for(int i=1;i<=n;i++){
    //    cout<<a[i];
//    }
    while(m--){//判断 
        int x;
        cin >> x;
        string s;
        cin >> s;
        if(s=="is"){
            string ss;
            cin >> ss;
            if(ss=="the"){
                string sss;
                cin >> sss;
                if(sss=="root"){
                    if(a[1]==x) cout << "T" << endl;
                    else cout << "F" <<endl;
                }
                else{
                    string s1;
                    int xx;
                    cin >> s1 >> xx;
                    if(find(x)==find(xx)/2) cout << "T" << endl;
                    else cout << "F" << endl;
                }
            }
            else if(ss=="a"){
                string s1,s2;
                int xx;
                cin >> s1 >> s2 >> xx;
                if(find(x)/2==find(xx)) cout << "T" << endl;
                else cout << "F" << endl;
            }
        }
        else if(s=="and"){
            int xx;
            string s1,s2;
            cin >> xx >> s1 >> s2;
            if(find(x)/2==find(xx)/2) cout << "T" << endl;
            else cout << "F" << endl;
        }
    }
    return 0;
} 
原文地址:https://www.cnblogs.com/liyongqi/p/14021935.html