1024. Palindromic Number (25)

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

类型：大数运算

题目大意：给定一个数字，和允许翻转后相加的次数cnt，求要多少次才能变成一个回文数字，输出那个回文数字和翻转相加了多少次，如果本身就是回文数字就输出0次，如果超过给定的次数cnt了，就输出那个不是回文的结果，并且输出给定的次数cnt

（1）没啥要说的，思路很清晰。主要的问题：就是在局部（非全局变量）申明一个字符串是默认初始化是随机的，且是不可以预测的。所以统一初始化为“”。这里我用的是char数组，没有用string容器，更可气的是G++.2.4.7编译器内没有strrev（char*）字符串翻转函数，就很伤。

#include
#include
using namespace std;
char Astr[150];
char Bstr[150];
int step;
void addstr(char *str1,char *str2)//字符串求和，结果给Astr 
{
	char sumstr[150]="";//局部字符串要初始化，否则默认初始化其他的值（二全局变量默认初始化位0） 
	char sumtemp[150]="";//之前定义字符串数组就是没有初始化，导致一些案例无法通过 
	int carry=0;
	int index=0;
	int last=0;
	for(int i=(int)strlen(str1)-1;i>=0;--i)
	{
		sumstr[index++]=(str1[i]+str2[i]-'0'-'0'+carry)%10+'0';
		carry=(str1[i]+str2[i]-'0'-'0'+carry)/10;
	}
	if(carry!=0)
		sumstr[index++]=carry+'0';
	for(int i=index-1;i>=0;--i)
		sumtemp[last++]=sumstr[i];
	strcpy(Astr,sumtemp);
}
bool judge(char *str)//判断字符串是否是回文字符串 
{
	int j=(int)strlen(str)-1;
	int i=0;
	while(i<=j)
	{
		if(str[i]!=str[j])
			return false;
		++i;
		--j;
	}
	return true;
}
void changestr(char *str)//Astr字符串倒置复制给Bstr 
{
	int j=0;
	char strtemp[150]="";
	for(int i=(int)strlen(str)-1;i>=0;--i)
		strtemp[j++]=str[i];
	//strtemp[j]='\0';//比不可少 ,如果定义字符串数组没有初始化的话，即char strtemp[150]; 
	strcpy(Bstr,strtemp);
}
int main()
{
	//freopen("in.txt","r",stdin);
	int realstep=0;
	cin>>Astr>>step;
	if(judge(Astr))
		cout<

（2） 分析：1.会超出long int类型（会有两个点溢出错误），所以用字符串存储，大整数相加
2.可以通过对字符串翻转后比较来判断是否为回文串

#include 
#include 
#include 
using namespace std;
string s;
void add() {
    string t = s;
    reverse(t.begin(), t.end());
    int len = s.length(), carry = 0;
    for(int i = 0; i < len; i++) {
        s[i] = s[i] + t[i] + carry - '0';
        carry = 0;
        if(s[i] > '9') {
            s[i] = s[i] - 10;
            carry = 1;
        }
    }
    if(carry) s += '1';
    reverse(s.begin(), s.end());
}
int main() {
    int cnt;
    cin >> s >> cnt;
    for(int i = 0; i <= cnt; i++) {
        string t = s;
        reverse(t.begin(), t.end());
        if(s == t) {
            cout << s << endl << i;
            return 0;
        }
        if(i != cnt) add();
    }
    cout << s << endl << cnt;
    return 0;
}