C38-解密宝箱

一、上周作业回顾

Hw01. A22299.下一个字母

常规简单题，注意类型的转换

#include<iostream>
using namespace std;
int main(){
    string a;
    cin>>a;
    for(int i=0;i<a.size();i++){
        if(a[i]>='0'&&a[i]<='9'){
            cout<<a[i];
        }
        else if(a[i]=='z'){
            cout<<'a';
        }
        else if(a[i]=='Z'){
            cout<<'A';
        }
        else{
            cout<<char(a[i]+1);
        }
    }
}

Hw02. A29344.50%AI, 50%Human

注意整数的除法自动向下取整以及输出格式。

#include <bits/stdc++.h>
using namespace std;

int main()
{
	int n, a=0, b=0;
	string s;
	cin >> n;
	for (int i=1; i<=n; i++){
		cin >> s;
		if (s.size() > 5) a++; 
	} 
	int x = ceil(a*1.0/n*100);
	int y = 100-x;
	cout<<x<<"%AI, "<<y<<"%Human"; 
	
	return 0;
}

二、课堂案例

/*
测试输入代码： 
.---- -.- --.- -.... .-.. ----- -.. .--- --- ....
*/
#include <bits/stdc++.h>
using namespace std;

string morse[37] = {".-", "-...", "-.-.", "-..", ".",
	                "..-.", "--.", "....", "..", ".---",
	                "-.-", ".-..", "--", "-.", "---",
	                ".--.", "--.-", ".-.", "...", "-",
	                "..-", "...-", ".--", "-..-", "-.--","--..",
					"-----",".----","..--","...--","....-",
					".....","-....","--...","---..","----."
					}; 	//存放摩斯密码的字符串数组 
					
string a[11];		//待解密的密文 
char b[11]; 		//1.摩斯密码的解密结果 
char c[11];			//2.凯撒解密的结果 
char d[3][6];		//3.栅栏解密的结果 

int idx; 

int main()
{	
	for (int i=0; i<10; ++i) cin>>a[i]; 	//输入密文 
	
    //1.摩尔斯解密
	//先 拿出 a , 将a里面的每一个与morse[j]进行比较 
	for (int i=0; i<10; i++) {
		for (int j=0; j<36; j++){
			if (a[i] == morse[j]){
			 	//morse[j]就是我要找的内容 
			 	if (j<26) b[idx++] = char(j+97);
			 	else b[idx++] = char(j+22);			//26+22 = 48 
			 	break; 
			}
		}
	}
	cout<< b << endl;

	//2.凯撒解密
	for (int i=0; i<10; i++){
		if (b[i]>='a' && b[i] <='c'){
			c[i] = b[i]  + 23;
		}
		else if (b[i]>='d' && b[i]<='z'){
			c[i] = b[i] - 3; 
		}
		else{
			c[i] = b[i];
		} 
	} 
	cout<< c << endl;
	
	
	//3.栅栏解密
	idx = 0;
	for (int  i=0; i<2;  i++){
		for (int j=0; j<5; j++){
			d[i][j] = c[idx++]; 
		}
	} 
	
	//输出栅栏解密结果
	for ( int i=0; i<5; i++){
		for (int j=0; j<2; j++){
			cout<<d[j][i];
		} 
	} 
	cout << endl;
	
	//4.杨辉三角 
	int e[25][25]= {0}, n=10;
	for(int i=1; i<=n; i++) {
		e[i][1]=1;
		e[i][i]=1;
	}
	for(int i=3; i<=n; i++) {
		for(int j=2; j<=i; j++) {
			e[i][j]=e[i-1][j]+e[i-1][j-1];
		}
	}
	cout << e[10][6] << endl; 

	return 0;
}