2026-06-25：给定范围内 K 位数字之和。用go语言，给定三个整数 l、r、k。 我们要考虑所有恰好有 k 位的整数，这些整数的每一位数字都是

package main

import (
    "fmt"
)

const mod = 1_000_000_007

func pow(x, n int)int {
    res := 1
    for ; n > 0; n /= 2 {
        if n%2 > 0 {
            res = res * x % mod
        }
        x = x * x % mod
    }
    return res
}

func sumOfNumbers(l, r, k int)int {
    m := r - l + 1
    return (l + r) * m * (pow(10, k) - 1 + mod) % mod * pow(18, mod-2) % mod * pow(m, k-1) % mod
}

func main() {
    l := 5
    r := 5
    k := 10
    result := sumOfNumbers(l, r, k)
    fmt.Println(result)
}

# -*-coding:utf-8-*-

MOD = 1_000_000_007

def pow_mod(x, n):
    """快速幂取模"""
    res = 1
    while n > 0:
        if n % 2 == 1:
            res = res * x % MOD
        x = x * x % MOD
        n //= 2
    return res

def sum_of_numbers(l, r, k):
    m = r - l + 1
    # (l + r) * m * (10^k - 1) * inv(18) * m^(k-1) mod MOD
    return (l + r) * m % MOD * (pow_mod(10, k) - 1 + MOD) % MOD * pow_mod(18, MOD - 2) % MOD * pow_mod(m, k - 1) % MOD

def main():
    l = 5
    r = 5
    k = 10
    result = sum_of_numbers(l, r, k)
    print(result)

if __name__ == "__main__":
    main()

#include <iostream>
using namespace std;

const long long MOD = 1000000007LL;

long long pow_mod(long long x, long long n) {
    long long res = 1;
    while (n > 0) {
        if (n % 2 == 1) {
            res = res * x % MOD;
        }
        x = x * x % MOD;
        n /= 2;
    }
    return res;
}

long long sumOfNumbers(long long l, long long r, long long k) {
    long long m = r - l + 1;
    long long result = (l + r) % MOD;
    result = result * m % MOD;
    result = result * ((pow_mod(10, k) - 1 + MOD) % MOD) % MOD;
    result = result * pow_mod(18, MOD - 2) % MOD;
    result = result * pow_mod(m, k - 1) % MOD;
    return result;
}

int main() {
    long long l = 5;
    long long r = 5;
    long long k = 10;
    long long result = sumOfNumbers(l, r, k);
    cout << result << endl;
    return0;
}