PAT 1136 A Delayed Palindrome[简单]
1136 A Delayed Palindrome (20 分)
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0≤ai<10for 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.
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!!!