Palindromic Numbers (III)

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

     

Description

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

Hint

Dataset is huge, use faster I/O methods.

AC CODE：

//Memory: 1608 KB		Time: 82 MS//Language: C++		Result: Accepted#include 
         
          #include 
          
           #include 
           
            using namespace std;char s[100001];int main(){    int T, c = 1;    int i, j, len, mid;    bool odd, flag;    scanf("%d", &T);    while(T--)    {        scanf("%s", s);        printf("Case %d: ", c++);        //特判所有digit为9的情况        for(i = 0; s[i] == '9'; i++){}        if(s[i] == '\0')        {            putchar('1');            for(i = 0; s[i+1] != '\0'; i++)                putchar('0');            puts("1");            continue;        }        len = strlen(s);        //特判只有一位数的情况        if(len < 2)        {            s[0]++;            puts(s);            continue;        }        mid = len / 2;        odd = len & 1;        if(odd)        {            for(i = mid - 1, j = mid + 1; i > -1 && s[i] == s[j]; i--, j++){}            //原数是回文数            if(i == -1)            {                for(i = mid; s[i] == '9'; i++)                    s[len - i - 1] = s[i] = '0';                s[len - i - 1] = ++s[i];                puts(s);            }            else if(s[i] > s[j])            {                for(i = 0; i < mid; i++)                    putchar(s[i]);                for(; i > -1; i--)                    putchar(s[i]);                puts("");            }            //s[i] < s[j]            else            {                /*for loop中的判断条件不需要i > -1,因为必定能找到s[i] != '9'的i。试想如果                s[i] == '9'一直成立，就不会有s[i] < s[j]了。*/                for(i = mid; s[i] == '9'; i--)                    s[i] = '0';                s[i]++;                for(i = 0; i < mid; i++)                    putchar(s[i]);                for(; i > -1; i--)                    putchar(s[i]);                puts("");            }        }        //odd = false        else        {            for(i = mid - 1, j = mid; i > -1 && s[i] == s[j]; i--, j++){}            //原数是回文数            if(i == -1)            {                /*此处for loop不需要i
            
              s[j])            {                for(i = 0; i < mid; i++)                    putchar(s[i]);                for(i = mid - 1; i > -1; i--)                    putchar(s[i]);                puts("");            }            //s[i] < s[j]            else            {                //必定存在i使得s[i]!='9',不然不会有s[i] < s[j](这是进入这个else的条件)                for(i = mid - 1; s[i] == '9'; i--)                    s[i] = '0';                s[i]++;                for(i = 0; i < mid; i++)                    putchar(s[i]);                for(i = mid - 1; i > -1; i--)                    putchar(s[i]);                puts("");            }        }    }}