bjfu1281

思路挺简单的，但因为需要处理大数，所以就比较耗代码了。

/*

 * Author    : ben

 */

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <string>

#include <vector>

#include <deque>

#include <list>

#include <functional>

#include <numeric>

#include <cctype>

using namespace std;



//比较两个数s1和s2的大小，当s1<s2时返回真

bool lessthan(const char* s1, const char* s2, int len) {

    for (int i = 0; i < len; i++) {

        if (s1[i] < s2[i]) {

            return true;

        } else if (s1[i] > s2[i]) {

            return false;

        }

    }

    return false;

}



//检测是否栅栏数

bool judge(const char* ss, int len) {

    if (len < 3 || len % 2 == 0) {

        return false;

    }

    int d = len / 2;

    for (int i = 0; i < d; i++) {

        if (ss[i] != '1' || ss[len - i - 1] != '1') {

            return false;

        }

    }

    return true;

}



int lowercount(const char* ss, int len) {

    if (len < 3) {

        return 0;

    }

    if (len % 2 == 0) {

        return (len / 2 - 1) * 10;

    }

    if (judge(ss, len)) {

        return (len / 2 - 1) * 10 + ss[len / 2] - '0';

    }

    char str[2000];

    memset(str, '1', len);

    str[len] = 0;

    str[len / 2] = '0';

    if (lessthan(ss, str, len)) {

        return (len / 2 - 1) * 10;

    } else {

        return (len / 2 ) * 10;

    }

}





int main() {

//    freopen("data.in", "r", stdin);

    char a[2000], b[2000];

    int T, lena, lenb;

    scanf("%d", &T);

    for (int t = 1; t <= T; t++) {

        scanf(" %s %s", a, b);

        lena = strlen(a);

        lenb = strlen(b);

        int ans = lowercount(b, lenb) - lowercount(a, lena);

        if (judge(b, lenb)) {

            ans++;

        }

        printf("%d\n", ans);

    }

    return 0;

}