10/01/20 The Next Palindrome 解题报告

5. The Next Palindrome

Problem code: PALIN

 

A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.

Input

The first line contains integer t, the number of test cases. Integers K are given in the next t lines.

Output

For each K, output the smallest palindrome larger than K.

Example

Input:
2
808
2133

Output:
818
2222
Warning: large Input/Output data, be careful with certain languages

 

#include <stdio.h> #include <stdlib.h> #include <string.h> #define DIGITS 1000000 //使用递归求解下一个回文数，原理是去除头尾的回文数，把子回文数递归求解 //有bug 例如19991的情况，而且递归内存申请无法求解大数据量 char* NextPalindrome(const char* k, int added) { char* buf; int len, mid, beg, end, i, carry; len = strlen(k); buf = (char*)malloc(len+2); strcpy(buf, k); if (len==1) { if (added) buf[0] = buf[0] + 1 ; else ; return buf; } if (buf[0]<buf[len-1]) { i = mid = (len-1)/2; beg = 0; end = len-1; carry = 0; buf[i--] = (carry=buf[i]-'0'+1)%10+'0'; while (i>=0&&carry>0)buf[i--] = (carry=buf[i]-'0'+carry/10)%10+'0'; if (carry/10==1) { for (i=len;i>=0;i--) { buf[i+1] = buf[i]; } buf[0] = '1'; buf[len+2] = '/0'; if(len%2==0)mid=mid+1; end = end+1; } while (end>mid) buf[end--] = buf[beg++]; return buf; } else if (buf[0]==buf[len-1]) { if (len>2) { char* part; buf[len-1] = '/0'; part = NextPalindrome(buf+1, 1); len = strlen(part); strcpy(buf+1, part); buf[len+1] = buf[0]; buf[len+2] = '/0'; } return buf; } else { if (len>2) { char* part; buf[len-1] = '/0'; part = NextPalindrome(buf+1, 0); len = strlen(part); strcpy(buf+1, part); buf[len+1] = buf[0]; buf[len+2] = '/0'; free(part); } return buf; } } int IsPalindrome(const char* k) { int len, mid, beg, end ; len = strlen(k); mid = (len-1)/2; beg = 0; end = len-1; while (end>mid) { if (k[beg++]!=k[end--]) return 0; } return 1; } void PalindromeInc(char* k) { int len, i, carry; len = strlen(k); i = len-1; carry = 0; k[i--] = (carry=k[i]-'0'+1)%10+'0'; while (i>=0&&carry>0)k[i--] = (carry=k[i]-'0'+carry/10)%10+'0'; if (carry/10==1) { for (i=len;i>=0;i--) { k[i+1] = k[i]; } k[0] = '1'; k[len+2] = '/0'; } } int main(int argc, char **argv) { int t; char K[DIGITS+2]; int len, mid, beg, end, i, carry, flag; scanf("%d", &t); // 读入测试用例数 while (t--) { scanf("%s", &K); // if (IsPalindrome(K)) // PalindromeInc(K); // printf("%s/n", NextPalindrome(K, 0)); len = strlen(K); // 整数长度 mid = (len-1)/2; // 取得回文数的中点，若是偶数则是前半段的最后一个字符,这个位置对于我们下一个回文数十分重要 beg = 0; end = len-1; carry = 0; // 进位 flag = 0; // 是否改变读入的数，更确切是否增加读入的数 // 启发模式1:检测，若前半段的字符大于后半段对应位置的字符，把前半段赋给后半段来增加数的大小 while (end>mid) { if (K[beg]>K[end]) { K[end] = K[beg]; flag = 1; } else if (K[beg]<K[end]) break; else ; end = end-1; beg = beg+1; } if (flag) // 如果应用启发模式1 检测当前数是否是回文数，若是则输出 { if (IsPalindrome(K)) { printf("%s/n", K); continue; } } // 对于非启发模式1的下一个回文数求解只需要把前半段增1,并镜射拷贝至后半段即可完成 i = mid; beg = 0; end = len-1; K[i--] = (carry=K[i]-'0'+1)%10+'0'; //自增算法 while (i>=0&&carry>0)K[i--] = (carry=K[i]-'0'+carry/10)%10+'0'; if (carry/10==1) // 是否为 99 ,9999, 999,之类需要增加串长的情形 { for (i=len;i>=0;i--) { K[i+1] = K[i]; } K[0] = '1'; K[len+2] = '/0'; end = end+1; len = len+1; mid = (len-1)/2; } while (end>mid) K[end--] = K[beg++]; //镜像拷贝 printf("%s/n", K); //输出结果 } return 0; }