1136 A Delayed Palindrome (20 分)
标签：until reset 内存限制 long sed round HERE ini bin
Consider a positive integer N written in standard notation with k+1 digits a?i?? 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.
Each input file contains one test case which gives a positive integer no more than 1000 digits.
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.
97152
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
196
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.
#include<bits/stdc++.h> using namespace std; typedef long long ll; string reversestr(string s){ string res = ""; for(auto x:s){ res = x+res; } return res; } string strplus(string a,string b){ string res = ""; int len = a.size(); int jinwei = 0; for(int i=len-1;i >= 0;i--){ int numa = a[i] - ‘0‘; int numb = b[i] - ‘0‘; int numsum = numa + numb + jinwei; jinwei = numsum/10; int add = numsum%10; res = to_string(add) + res; } if(jinwei) res = "1"+res; return res; } int main(){ string s; cin >> s; int cnt = 0; while(1){ string t = reversestr(s); if(t == s) { printf("%s is a palindromic number.", s.c_str()); break; } string kkp = strplus(s,t); printf("%s + %s = %s\n",s.c_str(),t.c_str(),kkp.c_str()); s = kkp; cnt++; if(cnt >= 10){ printf("Not found in 10 iterations."); break; } } return 0; }
大数相加突然变得好容易啊。。。
标签：until reset 内存限制 long sed round HERE ini bin
原文地址：https://www.cnblogs.com/cunyusup/p/10789547.html
