Another palindrome related problem. Actually nothing too theoretical here, but please keep following hints in mind:

1.　How to check whether a natural number is Palindrome

　　Not sure whether there's closed form to Palindrome, I simply used a naive algorithm to check: log10() to get number of digits, and check mirrored digits.

2.　Pre calculation

　　1<=a<=b<=1000. so we can precalculate all Palindromes within that range beforehand.

3.　Understand problem statement, only start from a Palindrome

　　For each range, it must start from a Palindrome - we can simply skip non-Palindromes. And don't forget to remove all tailing non-Palindromes.

 

// 692 Fruit Farm#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;/#define gc getchar_unlockedint read_int(){  char c = gc();  while(c<'0' || c>'9') c = gc();  int ret = 0;  while(c>='0' && c<='9') {    ret = 10 * ret + c - 48;    c = gc();  }  return ret;}int read_string(char *p){    int cnt = 0;    char c;    while((c = gc()) == ' ');    //    skip spaces    //    while(c != 10)    {        p[cnt ++] = c;        c = gc();    }    return cnt;}void print_fast(const char *p, int len){    fwrite(p, 1, len, stdout);}/bool isPalin(int n){    if(n >= 1 && n < 10) return true;    //    Get digit length    int nDigits = 1 + (int)floor(log10(n * 1.0));    //    Get separated digits    int digits[5] = {
    0};    for(int i = 0; i < nDigits; i ++)    {        int d = n / (int)pow(10.0, i*1.0) % 10;        digits[i] = d;    }    //    Check digits    bool bEven = nDigits % 2 == 0;    int inxLow = nDigits / 2 - 1;    int inxHigh = (nDigits / 2) + (bEven ? 0 : 1);    int nDigits2Check = nDigits / 2;    for(int i = 0; i < nDigits2Check; i ++)    {        if(digits[inxLow] != digits[inxHigh]) return false;        inxLow --; inxHigh ++;    }    return true;}bool Palin[1000] = {
    false};void precalc_palin(){    for(int i = 1; i <= 1000; i ++)    {        if(isPalin(i))        {            Palin[i-1] = true;            //printf("%d ", i);        }    }    //printf("\n");}void calc(int a, int b, int l){    int rcnt = 0; int mya = 0, myb = 0;    for(int i = a; i <= b; i++)    {        if(!Palin[i-1]) continue;        //printf("At %d\n", i);        int cnt = 0; int bound = min(b, i + l - 1);        for(int j = i; j <= bound; j ++)        {            if(Palin[j-1])    cnt ++;        }        //printf("[%d, %d] = %d\t", i, bound, cnt);        if(cnt > rcnt)        {            rcnt = cnt; mya = i; myb = bound;        }    }    // shrink    if(rcnt > 0)    {        while(!Palin[myb-1]) myb--;        printf("%d %d\n", mya, myb);    }    else    {        printf("Barren Land.\n");    }}int main(){    //    pre-calc all palindrome in [1-1000]    precalc_palin();    int runcnt = read_int();    while(runcnt--)    {        int a = read_int();        int b = read_int();        int l = read_int();        calc(a, b, l);    }    return 0;}