PE 145 How many reversible numbers are there below one-billion? （暴力）-蒲公英云

PE 145 How many reversible numbers are there below one-billion? （暴力）

How many reversible numbers are there below one-billion?

Problem 145

Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).

There are 120 reversible numbers below one-thousand.

How many reversible numbers are there below one-billion (109)?

题意：

有多少小于十亿的可逆数？

有些正整数n满足这样一种性质，将它的数字逆序排列后和本身相加所得到的和[ n + reverse(n) ]的十进制表示只包含有奇数数字。例如，36 + 63 = 99 以及409 + 904 = 1313。我们称这样的数是可逆的；因此36，63，409和904都是可逆的。无论是n还是reverse(n)都不允许出现前导0。

在小于一千的数中，一共有120个可逆数。

在小于十亿（109）的数中，一共有多少个可逆数？

题解：感觉没什么好说的….直接暴力….

其实一亿到10亿之间是没有可逆数的。然后暴力到一亿就可以了。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int reverse(ll x)
{
    ll r=0;
    while (x>0)
    {
        r=r*10+(x%10);
        x/=10;
    }
    return r;
}
int check_odd(ll x)
{
    int flag=0;
    while (x>0)
    {
        if ((x%10)%2==0){
            flag=1;
            break;
        }
        x/=10;
    }
    if (flag) return 0;
    else return 1;
}
int main()
{
    ll count=0;
    //在 1 亿到 10亿之间没有可逆数 
    for (ll i=10;i<=100000000;i+=2)
    {
        if (check_odd(i+reverse(i)) && i%10!=0) 
        count++;
    } 
    cout<<count*2<<endl;
    return 0;
}