#include <iostream> using namespace std; int Loop(int x) { int sum = 0, n = x; while(n > 0) { sum = sum * 10 + n % 10; n = n / 10; } if(x == sum) return 1; else return 0; } int main() { int a, b, i, n; do{ cin >> a >> b; }while(a >= b); for(i = a; i <= b; i ++) { n = Loop(i); if(i % 2 == 0 && n == 1) cout << i << endl; } return 0; }增加这段代码的范围
时间: 2023-12-18 17:03:59 浏览: 68
这段代码是用来判断一个区间内的所有偶数是否为回文数,并输出回文数的代码。
如果要增加这段代码的范围,可以修改输入部分的代码,如下所示:
```cpp
int main() {
int a, b, i, n;
cout << "请输入区间的起始数和结束数:" << endl;
cin >> a >> b;
while(a >= b) {
cout << "起始数必须小于结束数,请重新输入:" << endl;
cin >> a >> b;
}
for(i = a; i <= b; i ++) {
n = Loop(i);
if(i % 2 == 0 && n == 1)
cout << i << endl;
}
return 0;
}
```
这样修改后,用户需要输入起始数和结束数,如果起始数大于等于结束数,则会提示用户重新输入。这样就增加了代码的范围,使得用户可以输入任意区间进行判断。
相关问题
int main() { int n,m,i,sum=0; cin>>n>>m; for( i=49;i<=m;i++) { sum+=i%10; i=i/10; sum=sum+i; if(sum==13) cout<<i<<endl; } return 0; }
This program reads two integers n and m from the user and then loops through all integers from 49 to m.
For each integer i in the loop, the program calculates the sum of its digits using the modulo operator and integer division. This sum is stored in the variable sum.
Then, the program divides i by 10 and adds the result to the sum variable. If the resulting sum is equal to 13, the program prints i to the console.
Note that there is a syntax error in the program: the loop condition should be i<=m instead of i to avoid an infinite loop. Additionally, there is no need to declare the variable sum outside of the loop.
Here is a corrected version of the program:
```cpp
#include <iostream>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
for (int i = 49; i <= m; i++) {
int sum = 0;
int temp = i;
while (temp > 0) {
sum += temp % 10;
temp /= 10;
}
sum += i / 10;
if (sum == 13) {
cout << i << endl;
}
}
return 0;
}
```
This version of the program correctly calculates the sum of digits of each integer using a while loop and a temporary variable. It also declares the sum variable inside the loop to avoid reusing its value from previous iterations.
阅读全文