福尔认为，如果一个数x 中各位数之和为一个设定的值y ，则这个数是一个幸运数。福尔想知道的是，给定一个整数a ，不小于a 的最小幸运数是什么？ 输入数据 输入数据有多组，每组为两个数串a,y(0≤a≤101000,0≤y≤105) ，分别为如上所述的两个整数。 输出数据 对每组输入数据，请在单独的行中数串满足条件的最小的幸运数，若不存在则输出-1。写c++代码

以下是一个可行的C++代码实现，使用了字符串类 `string` 和高精度运算类 `BigInt`：

#include <iostream>
#include <string>
using namespace std;

// 高精度整数类
class BigInt {
public:
    BigInt() {}
    BigInt(const string& s) : num(s) {}
    BigInt operator+(const BigInt& b) const;
    BigInt operator-(const BigInt& b) const;
    BigInt operator*(int n) const;
    bool operator<(const BigInt& b) const { return num < b.num; }
    bool operator<=(const BigInt& b) const { return num <= b.num; }
    bool operator>(const BigInt& b) const { return num > b.num; }
    bool operator>=(const BigInt& b) const { return num >= b.num; }
    bool operator==(const BigInt& b) const { return num == b.num; }
    bool operator!=(const BigInt& b) const { return num != b.num; }
    friend ostream& operator<<(ostream& out, const BigInt& x); // 输出流重载
    string num;
};

// 重载加法运算
BigInt BigInt::operator+(const BigInt& b) const {
    string a = num;
    string c = b.num;
    int n1 = a.size();
    int n2 = c.size();
    if (n1 < n2) {
        for (int i = 0; i < n2 - n1; i++) {
            a = "0" + a;
        }
    } else {
        for (int i = 0; i < n1 - n2; i++) {
            c = "0" + c;
        }
    }
    int carry = 0;
    string res = "";
    for (int i = a.size() - 1; i >= 0; i--) {
        int sum = a[i] - '0' + c[i] - '0' + carry;
        res = to_string(sum % 10) + res;
        carry = sum / 10;
    }
    if (carry) {
        res = "1" + res;
    }
    return BigInt(res);
}

// 重载减法运算
BigInt BigInt::operator-(const BigInt& b) const {
    string a = num;
    string c = b.num;
    int n1 = a.size();
    int n2 = c.size();
    if (n1 < n2) {
        for (int i = 0; i < n2 - n1; i++) {
            a = "0" + a;
        }
    } else {
        for (int i = 0; i < n1 - n2; i++) {
            c = "0" + c;
        }
    }
    int borrow = 0;
    string res = "";
    for (int i = a.size() - 1; i >= 0; i--) {
        int diff = a[i] - '0' - (c[i] - '0') - borrow;
        if (diff < 0) {
            diff += 10;
            borrow = 1;
        } else {
            borrow = 0;
        }
        res = to_string(diff) + res;
    }
    while (res.size() > 1 && res[0] == '0') {
        res.erase(0, 1);
    }
    return BigInt(res);
}

// 重载乘法运算
BigInt BigInt::operator*(int n) const {
    int carry = 0;
    string res = "";
    for (int i = num.size() - 1; i >= 0; i--) {
        int product = (num[i] - '0') * n + carry;
        res = to_string(product % 10) + res;
        carry = product / 10;
    }
    if (carry) {
        res = to_string(carry) + res;
    }
    return BigInt(res);
}

// 重载输出流
ostream& operator<<(ostream& out, const BigInt& x) {
    out << x.num;
    return out;
}

// 求字符串数字的各位数字之和
int digitSum(const string& s) {
    int sum = 0;
    for (int i = 0; i < s.size(); i++) {
        sum += s[i] - '0';
    }
    return sum;
}

int main() {
    string a, y;
    while (cin >> a >> y) {
        BigInt A(a), Y(y);
        BigInt ans = A; // 初始值为a
        while (digitSum(ans.num) != Y.num[0] - '0') {
            ans = ans + BigInt("1");
        }
        if (ans < A) {
            cout << "-1" << endl;
        } else {
            cout << ans << endl;
        }
    }
    return 0;
}

该程序的核心部分是求最小幸运数的循环部分，每次将当前数加1，直到各位数字之和等于设定的值。如果最终结果小于原数a，则说明不存在符合条件的幸运数，输出-1。