大于N的最小回文数

题目描述

Palindromic numbers are digital strings that read the same both forwards and backwards. For example, 121, 44 and 3 are Palindromic numbers, 175, 36 are not;For a given integer N, you are to find a palindromic number P that satisfy P>N. However, there are many such palindromic numbers. Your task is to find the least one.

输入

There are several test cases, each test case contains only one positive integer N in one line. The number of digits of N is not exceeding 10,000, and N has not lead zeroes.The input will finish with the end of file.

输出

For each the case, your program will output the least palindromic number P (P > N) on separate line.

样例输入

443175

样例输出

554181

来源

2011ACM/ICPC 第五届华中南赛区邀请赛 

题解

思路

题意要求，数N可达10000位，只能用字符串数组存储，对于回文数，只要求前半部分就可写出后半部分，大致分两种情况，一种是前半部分小于后半部分，前半部分有要+1的，这时候就要判断是否有9，有9存在要增加一位的情况，如999/199/191。另一种简单，因为前半部分大于后半部分，所以，只要把前半部分替换后半部分即可，如321/291.

测试数据

89999
1231
12351
9998
9898
2993
1000
100
192
291
121
99
999
2
10
1991
19991
19191

代码如下

#include <cstdio>#include <cstring>#define N 10001char s[N];using namespace std;int main(){    while ( ~scanf("%s",s)){        int len=strlen(s);        int flag=0,i;        for(i=len/2-1;i>=0;--i){            if(s[i]>s[len-1-i]){flag=1;break;}            else if(s[i]<s[len-1-i]){ flag=-1;break;}        }        if(flag!=1){//前半串要加1            //s[(len-1)/2]++;            for(i=(len-1)/2;i>=0;--i){//199 191 999                s[i]++;                if(s[i]>'9'){                    s[i]='0';                }else break;            }            if(s[0]=='0'){//999 9999                s[0]='1';                len++;                s[len/2]='0';            }        }        for(i=0;i<len/2;++i)            printf("%c",s[i]);        for(i=(len+1)/2-1;i>=0;--i)            printf("%c",s[i]);        printf("\n");    }    return 0;}