1136 A Delayed Palindrome (20 point(s))
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0≤ai<10 for all i and ak>0. Then N is palindromic if and only if ai=ak−i for all i. Zero is written 0 and is also palindromic by definition.
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. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )
Given any positive integer, you are supposed to find its paired palindromic number.
Input Specification:
Each input file contains one test case which gives a positive integer no more than 1000 digits.
Output Specification:
For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:
A + B = C
where A
is the original number, B
is the reversed A
, and C
is their sum. A
starts being the input number, and this process ends until C
becomes a palindromic number -- in this case we print in the last line C is a palindromic number.
; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations.
instead.
Sample Input 1:
97152
Sample Output 1:
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
Sample Input 2:
196
Sample Output 2:
196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.
经验总结:
emmmm 题意是,给你一个数,让你判断它是否是回文数,如果不是,进行10次转换,看能否转为回文数,转换方法是,原回文数加上逆置的原回文数,判断相加的和是否是回文数。注意数的长度可能达到1000,利用大整数加法可以轻松解决~
AC代码
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=1010;
int n,m,k,num;
char s[maxn];
struct bign
{
int len,d[maxn];
bign()
{
len=0;
memset(d,0,sizeof(d));
}
};
bool isparlindrome(bign a)
{
for(int i=0;i<a.len/2;++i)
if(a.d[i]!=a.d[a.len-1-i])
return false;
return true;
}
bign change(char str[])
{
bign a;
a.len=strlen(str);
for(int i=0;i<a.len;++i)
a.d[i]=str[a.len-1-i]-'0';
return a;
}
void print(bign a)
{
for(int i=a.len-1;i>=0;--i)
printf("%d",a.d[i]);
}
bign invert(bign a)
{
bign b;
b.len=a.len;
for(int i=0;i<a.len;++i)
b.d[i]=a.d[a.len-1-i];
while(b.d[b.len-1]==0&&b.len>1)
--b.len;
return b;
}
bign add(bign a,bign b)
{
bign c;
c.len=max(a.len,b.len);
int carry=0;
for(int i=0;i<c.len;++i)
{
int temp=a.d[i]+b.d[i]+carry;
c.d[i]=temp%10;
carry=temp/10;
}
if(carry!=0)
c.d[c.len++]=carry;
return c;
}
int main()
{
scanf("%s",s);
bool flag=false;
bign c=change(s);
for(int i=0;i<=10;++i)
{
if(isparlindrome(c))
{
flag=true;
break;
}
if(i==10)
break;
bign a=c;
bign b=invert(c);
print(a);printf(" + ");
print(b);printf(" = ");
c=add(a,b);
print(c);printf("\n");
}
if(flag)
{
print(c);
printf(" is a palindromic number.\n");
}
else
printf("Not found in 10 iterations.\n");
return 0;
}