用字符串存储方便反转

解题思路:

注意事项:

参考代码:

#include<bits/stdc++.h>

using namespace std;

int init(string s)

{

int x = 0;

while (1) {

if (x >= 8)

{

return 0;

}

string s2 = s, s3;

int s1 = s.size();

reverse(s2.begin(), s2.end());

int n = 0;

for (int i = 0; i < s1; i++)

{

if (i == 0)

{

n = (s[i] - '0') + (s2[i] - '0');

}

else {

n += ((((s[i] - '0') + (s2[i] - '0'))) * pow(10, i));

}

}

while (n)

{

int p = n % 10;

n /= 10;

s3 += p + '0';

}

reverse(s3.begin(), s3.end());

bool flag = false;

for (int i = 0, j = s3.size() - 1; i <= j; i++, j--)

{

if (s3[i] != s3[j])

{

flag = true;

break;

}

}

if (!flag) {

return x + 1;

}

s = s3;

x++;

}

}

int main()

{

int T;

cin >> T;

while (T--)

{

string s;

cin >> s;

int ans=init(s);

cout << ans << endl;

}

}