1136 A Delayed Palindrome （20 分）
Consider a positive integer N written in standard notation with k+1 digits a?i?? as a?k???a?1??a?0?? with 0≤a?i??<10for 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.题目大意：给出一个数，是否回文？否则将其反转然后和原数相加，直到加到第10次未出现或者出现。
#include <iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstdlib>
using namespace std;
//但是属于哪一条地铁，怎么去表示呢？
bool pd(string s){
int sz=s.size();
for(int i=0;i<sz/2;i++){
if(s[i]!=s[sz-i-1])
return false;
}
return true;
}
int toNum(string s){
int a=0;
for(int i=0;i<s.size();i++){
a=a*10+(s[i]-‘0‘);
}
return a;
}
string toStr(int a){
string s;
while(a!=0){
s+=(a%10+‘0‘);
a/=10;
}
reverse(s.begin(),s.end());
return s;
}
int main(){
string s;
cin>>s;
string t=s;
//reverse(s[0],s[0]+s.size());//那个反转函数是啥来着???
//s.reverse();//这个调用不正确
reverse(s.begin(),s.end());//t是第一个数，s是第二个数.
bool flag=false;
int a,b,c;
for(int i=0;i<10;i++){
if(pd(s)){//如果这个数本来就是答案。
flag=true;break;
}
a=toNum(t);
b=toNum(s);
c=a+b;
cout<<t<<" + "<<s<<" = "<<c<<‘\n‘;
s=toStr(c);
t=s;
reverse(s.begin(),s.end());
}
if(flag){
cout<<t<<" is a palindromic number.";
}else{
cout<<"Not found in 10 iterations.";
}
return 0;
}
//本来是这么写的，但是提交只有18分，最后一个测试点过不去：答案错误。
//我明白了，这个的考点是大数相加(1K位)，而我却将其转换成int，实在是太不好了。
#include <iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstdlib>
using namespace std;
string add(string a){
string b=a;
reverse(b.begin(),b.end());
int jin=0,s;
for(int i=a.size()-1;i>=0;i--){
s=(a[i]-‘0‘)+(b[i]-‘0‘)+jin;
if(s>=10){
s%=10;
jin=1;
}else jin=0;
a[i]=s+‘0‘;
}
if(jin==1){
a="1"+a;
}
return a;
}
int main(){
string s;
cin>>s;
string t=s;
//reverse(s[0],s[0]+s.size());//那个反转函数是啥来着???
//s.reverse();//这个调用不正确
reverse(s.begin(),s.end());//t是第一个数，s是第二个数.
bool flag=false;
for(int i=0;i<10;i++){
if(t==s){//如果这个数本来就是答案。
flag=true;break;
}
cout<<t<<" + "<<s<<" = "<<add(t)<<‘\n‘;
t=add(t);
s=t;
reverse(s.begin(),s.end());
}
if(flag){
cout<<t<<" is a palindromic number.";
}else{
cout<<"Not found in 10 iterations.";
}
return 0;
}
//终于正确了，这水题花了我好长时间，哭唧唧！你看清题好吗？1000位的数字就是模拟大数相加啊！！！
原文地址：https://www.cnblogs.com/BlueBlueSea/p/9889496.html