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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??<10 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.
1 #include <map> 2 #include <set> 3 #include <queue> 4 #include <cmath> 5 #include <stack> 6 #include <vector> 7 #include <string> 8 #include <cstdio> 9 #include <cstring> 10 #include <climits> 11 #include <iostream> 12 #include <algorithm> 13 #define wzf ((1 + sqrt(5.0)) / 2.0) 14 #define INF 0x3f3f3f3f 15 #define LL long long 16 using namespace std; 17 18 const int MAXN = 2e3 + 10; 19 20 char A[MAXN], B[MAXN], C[MAXN]; 21 22 void calcB() 23 { 24 int len = strlen(A), a = len - 1, b = 0; 25 for (a ,b ; b < len; ++ b, -- a) 26 B[b] = A[a]; 27 } 28 29 void calcC() 30 { 31 int len1 = strlen(A), len = len1, b = 0; 32 int temp[MAXN]; 33 for (int i = 0, j = len1 - 1; i < len; ++ i, -- j) 34 { 35 if (j != -1) b += int(A[j] - ‘0‘) + int(B[j] - ‘0‘); 36 temp[i] = b % 10; 37 b /= 10; 38 if (i == len - 1 && b > 0) ++ len; 39 } 40 41 for (int i = 0, j = len - 1; i < len; ++ i, -- j) 42 C[i] = char (‘0‘ + temp[j]); 43 } 44 45 int main() 46 { 47 scanf("%s", &A); 48 int len = strlen(A), a = len - 1, b = 0; 49 for (a ,b ; b < len; ++ b, -- a) 50 B[b] = A[a]; 51 for (int i = 0; ; ++ i) 52 { 53 if (i == 10) 54 { 55 printf("Not found in 10 iterations.\n"); 56 break; 57 } 58 calcB(); 59 if (strcmp(A, B) == 0) 60 { 61 printf("%s is a palindromic number.\n", A); 62 break; 63 } 64 calcC(); 65 printf("%s + %s = %s\n", A, B, C); 66 strcpy(A, C); 67 } 68 return 0; 69 }
pat 1136 A Delayed Palindrome（20 分）
标签：any text https tps char mes org roc first
原文地址：https://www.cnblogs.com/GetcharZp/p/9580080.html
