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October 20 2025
//6-1-1 if
#include <iostream>
using namespace std;
int main() {
/*if (express rule)
rule1
*/
int a = 43545435341;
if (a % 2 == 1){
cout << “a is an odd number” << endl;
}
int b;
cin >> b;
//==1 can be ignore, because if can recognize the 0(False) or the other numbers(True)
if (b % 2 ) {
cout << “b is an odd number” << endl;
}
if (b % 2)
cout << “b is an odd number” << endl;// if there is no {} symbol to select, there won’t execute the next line in this if express
b = 3;// this one won’t be executed like the upper lineWW
cout << “b is ” << b << endl;
return 0;
}
//6-1-2 if else
#include <iostream>
using namespace std;
int main() {
/*
if (exp1){
rule1
}
else{
rule2
}
*/
int x;
cin >> x;
string y = “add brick”;
if (x & 1) {
cout << “x is an odd number” << endl;
y = y + ” JAVA”;
}
else {
cout << “x is an even number” << endl;
}
cout << y << endl;
return 0;
}
AI：
这个问题抓得很准，位与判断奇数的核心就藏在二进制的最末位里！
位与（&）能判断奇数，本质是利用了二进制数的末尾特征和位与运算的规则。当一个数和数字 1 进行位与时，结果直接反映了这个数的奇偶性。
1. 关键前提：二进制的 “奇偶密码”
所有整数的二进制表示中，最后一位（最低位）直接决定了它是奇数还是偶数：
- 奇数：二进制末尾一定是 1。比如 3（11）、5（101）、7（111）。
- 偶数：二进制末尾一定是 0。比如 2（10）、4（100）、6（110）。
这是因为二进制中，末尾位的权重是 \(2^0=1\)，其他位的权重都是 2 的倍数（如 \(2^1=2\)、\(2^2=4\)）。一个数的奇偶，本质就是看它是否包含 “1” 这个最小的奇数分量。
2. 核心规则：位与（&）的运算逻辑
位与运算的规则很简单：两个二进制位必须同时为 1，结果才是 1；否则结果为 0。
当我们用任意整数 n 和数字 1 进行位与时（即 n & 1），实际只关注 n 的二进制最后一位：
- 若 n 是奇数（末尾为 1）：1 & 1 = 1，最终结果为 1。
- 若 n 是偶数（末尾为 0）：0 & 1 = 0，最终结果为 0。
3. 实例验证：一看就懂
用具体数字举例，能更直观理解：
| 十进制数（n） | 二进制表示 | n & 1 的运算 | 运算结果 | 判断结论 |
| 3（奇数） | 11 | 11 & 01 = 01 | 1 | 奇数 |
| 4（偶数） | 100 | 100 & 001 = 000 | 0 | 偶数 |
| 7（奇数） | 111 | 111 & 001 = 001 | 1 | 奇数 |
| 8（偶数） | 1000 | 1000 & 0001 = 0000 | 0 | 偶数 |
要不要我帮你整理一份位运算判断奇偶的代码示例表？包含 Python、Java 等常见语言的具体写法，你可以直接参考使用。
//6 – 1 – 3 else if
#include <iostream>
using namespace std;
int main() {
int a;
cin >> a;
if (a==0) {
cout << “Vito Algorithm” << endl;
}
//if (如果)
//else if(否则如果)
else if (a==1) {
cout << “Vito1 Algorithm Legendary” << endl;
}
else if (a == 2) {
cout << “Vito Algorithm Killer” << endl;
}
else if (a == 3) {
cout << “Vito Algorithm Learner” << endl;
}
else{
//the last else can be ignored
cout << “Vito Another Algorithm” << endl;
}
return 0;
}
//6 – 1 – 4 conditional/ternary operator
#include <iostream>
using namespace std;
int main() {
int a = 3, b = 4;
float c = 1.6;
int x = -1;
if (a > b) {
x = a;
}
else {
x = b;
}
//exp1? exp2:exp3;
//if expression1 is true, then execute expression2, otherwise execute expresssion3
x = (a > b ? a : b);
x = (a > b) ? ( (a>c) ? a : c) : b; //ternary operator supports nest
// the value type in ternary is depend on the values’ type
cout << ((a > b) ? ((a > c) ? a : c) : c)<<endl;//1.6,output a float number
cout << ((a < b) ? ((a > c) ? a : c) : c) << endl;//3,output an int number
//If you think you can’t solve the priority problems, then you better add some quote symbols.
return 0;
}
//6 – 1 – 5 switch case
#include <iostream>
using namespace std;
int main() {
int a = 0;
cin >> a;
/*
if (a == 0) {
cout << “Zero” << endl;
}
else if(a == 1) {
cout << “One” << endl;
}
else if (a == 2) {
cout << “Two” << endl;
}
else if (a == 3) {
cout << “Three” << endl;
}
else{
cout << “Beyond three or below zero” << endl;
}
*/
switch (a) {
case 0:
cout << “Zero” << endl;
break; // If we didn’t add this break, then would execute next line until met the break, which is execute from up to down
case 1:
cout << “One” << endl;
case 2:
cout << “Two” << endl;
break;
case 3:
cout << “Three” << endl;
break;
/*case 7.7:
cout << “Three” << endl;
break;*/
case ‘4’:
cout << “Four” << endl;
break;
//the switch expression must use int number or the ” char to set the case, but the char ” will be recognized as the ascii codes
case ‘9’-‘2’:
//offset (偏移量),’9′ is 57 , ‘2’ is 50 , they are minus each other will be 7
cout << “‘9’-‘2’ is 7” << endl;
break;
default:
//If there weren’t meet the condition, if would jumped to the default case, which is equal to the if-else grammar’s else condition
cout << “Beyond three or below zero” << endl;
}
return 0;
}