Vinci is a little boy and is very creative. One day his teacher asked him to write all the Palindromic numbers from 1 to 1000. He became very frustrated because there is nothing creative in the task. Observing his expression, the teacher replied, "All right then, you want hard stuff, you got it." Then he asks Vinci to write a palindromic number which is greater than the given number. A number is called palindromic when its digits are same from both sides. For example: 1223221, 121, 232 are palindromic numbers but 122, 211, 332 are not. As there can be multiple solutions, Vinci has to find the number which is as small as possible.
Input
Input starts with an integer T (≤ 30), denoting the number of test cases.
Each case starts with a line containing a positive integer. This integer can be huge and can contain up to 105 digits.
Output
For each case, print the case number and the minimum possible palindromic number which is greater than the given number.
Sample Input
5
121
1
1332331
11
1121
Sample Output
Case 1: 131
Case 2: 2
Case 3: 1333331
Case 4: 22
Case 5: 1221
题意：输出比X大的第一个回文串。
思路：做过很多次了感觉，还是WA了几发。 按照如下几个步骤。
1，我们先按照左边几位对称到右边，如果比原串大，输出。
2，从中间到左边找第一个非‘9’的字符，+1，中间的全部变为‘0’。
3，全部都是‘9’，则长度+1，首尾为‘1’，其他为‘0’；
#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
char d[maxn],a[maxn];int cnt;
void put(int len)
{
if(len==-){
putchar(''); rep(i,,cnt-) putchar('');
putchar(''); putchar('\n'); return ;
}
rep(i,,len) putchar(a[i]); putchar('\n');
}
void cal()
{
cnt=strlen(d+);
rep(i,,cnt/) swap(d[i],d[cnt+-i]);
rep(i,cnt/+,cnt) a[i]=d[i];
rep(i,,cnt/) a[i]=a[cnt+-i];
for(int i=cnt;i>=;i--)
if(a[i]>d[i]){
put(cnt); return ;
}
else if(a[i]<d[i]) break;
rep(i,cnt/+,cnt){
if(a[i]<'') {
a[i]++;
rep(j,cnt-i+,i-) a[j]='';
a[cnt+-i]=a[i];
put(cnt); return ;
}
}
put(-); return ;
}
int main()
{
int T,C=,x;
scanf("%d",&T);
while(T--){
scanf("%s",d+);
printf("Case %d: ",++C);
cal();
}
return ;
}