PAT 甲级 1024. Palindromic Number
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1024. Palindromic Number (25)
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 3Sample Output 1:
4842Sample Input 2:
69 3Sample Output 2:
13533
题目大意:给你一个n和一个k,这个n反转以后与原来的数相加能否得到一个在K步内得到一个回文数,如果能就输出步数和数字,不能的话就输出k和数字
这道题好坑.....原来如果输入的数字本身就是回文数的话,他的步数是0.......
#include<iostream>#include<string>#include<algorithm>#include<cstdio>using namespace std;string add(string &a,string &b){//cout<<"*******"<<a<<' '<<b<<endl;string tmp;int carry=0;for(int i=a.size()-1;i>=0;i--){tmp.insert(tmp.begin(),((a[i]-'0')+(b[i]-'0')+carry)%10+'0');carry=((a[i]-'0')+(b[i]-'0')+carry)/10;}if(carry) tmp.insert(tmp.begin(),carry+'0');return tmp;}int main(){//freopen("in.txt","r",stdin);string str,tmp;int step;cin>>str>>step;int i;for(i=0;i<step;i++){tmp=str;reverse(tmp.begin(),tmp.end());if(tmp==str){cout<<str<<endl<<i<<endl;break;}else{str=add(str,tmp);}}if(i>=step) cout<<str<<endl<<step<<endl;}
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