pat 1136
1136 A Delayed Palindrome (20分)
Consider a positive integer N written in standard notation with k+1 digits ai?? as a?k???a?1??a?0?? with 0 for all i and a?k??>0. Then N is palindromic if and only if a?i??=a?k?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.
题意:给定一个数,判断是否为水仙花数,若不是,用这个数和它的逆置数的和取代它继续判断,10次取代之后如果扔不是,结束程序
思路:algorithm数组下的reverse(s.begin(),s.end())函数可以逆置字符串。
代码瑞如下:
#include<cstdio> #include<iostream> #include<algorithm> #include<string> using namespace std; string rev(string s){ reverse(s.begin(),s.end()); return s; } string add(string s1,string s2){ int count=0; int len=s1.length(); string s=s1; for(int i=len-1;i>=0;i--){ s[i]=(s1[i]-'0'+s2[i]-'0'+count)%10+'0'; count=(s1[i]-'0'+s2[i]-'0'+count)/10; } if(count==1) s="1"+s; return s; } int main(){ string s,sum; cin>>s; if(s==rev(s)){ cout<<s<<" is a palindromic number." ; return 0; } int mark=0; for(int i=0;i<10;i++){ sum=add(s,rev(s)); cout<<s<<" + "<<rev(s)<<" = "<<sum<<endl; s=sum; if(s==rev(s)){ cout<<s<<" is a palindromic number." ; mark=1; break; } } if(mark==0){ cout<<"Not found in 10 iterations."; } return 0; }