PAT甲级1024考试有哪些备考建议？

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1024. 回文数（25分） 时间限制：400ms 内存限制：65536kB 代码长度限制：16000B 题目程序：Standard 作者：CHEN, Yue

回文数是指正着读和反着读都一样的数。例如，12321 和 88 都是回文数。

请编写一个程序，判断给定的整数是否为回文数。

1024. Palindromic Number (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (<= 10¹⁰) is the initial numer and K (<= 100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484 2

Sample Input 2:

69 3

Sample Output 2:

1353 3

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
bool checkPalindromic(deque<char>dq)
{
int i = 0, j = dq.size() - 1;
while (i <= j)
{
if (dq[i] != dq[j])
return false;
i++; j--;
}
return true;
}
int main()
{
char N[20];
int K;
scanf("%s %d", N, &K);
int len = strlen(N);
deque<char> dq,dq1,dq2;
for (int i = 0; i < len; i++)
{
dq.push_back(N[i]);
}
dq1 = dq2 = dq;
reverse(dq2.begin(), dq2.end());
int i = 1;
int t, carry = 0;
if (checkPalindromic(dq))
{
for (int i = 0; i < dq.size(); i++)
{
printf("%c", dq[dq.size() - 1 - i]);
}
printf("\n%d", 0);
return 0;
}//若本身就是回文数，则不需进行计算了
while (i <=K)
{
for (int j = 0; j < dq.size(); j++)
{
t = dq1[j] - '0' + dq2[j] - '0' + carry;
carry = 0;
if (t >= 10)
{
dq[j] = '0' + t - 10;
carry = 1;
}
else
dq[j] = '0' + t;
}
if (carry)
{
dq.push_back('1');
carry = 0;
}
if (checkPalindromic(dq))
{
break;
}
else
{
dq1 = dq2 = dq;
reverse(dq2.begin(), dq2.end());
i++;
}
}
if (i > K)
{
for (int i = 0; i < dq.size(); i++)
{
printf("%c", dq[dq.size() - 1 - i]);
}
printf("\n%d",K);
}
else
{
for (int i = 0; i < dq.size(); i++)
{
printf("%c", dq[dq.size() - 1 - i]);
}
printf("\n%d", i);
}
return 0;
}