Leetcode每日一题 —— 3871. 统计范围内的逗号 II

class Solution:

    def countCommas(self, n: int) -> int:

        if n < 1000: return 0

        x = 1000

        ans = 0

        while x <= n:

            ans += n - x + 1

            x *= 1000

        return ans

无他，分段处理即可。

class Solution {

public:

    long long countCommas(long long n) {

        // 分段处理即可

        return max(0LL,n-1000+1)+max(0LL,n-1000000+1)+max(0LL,n-1000000000+1)+max(0LL,n-(long long)(1e12)+1)+max(0LL,n-(long long)(1e15)+1);

    }

};

最慢的一集

class Solution {

public:

    long long countCommas(long long n) {

        long long res = 0;

        int base = 0;

        long long val = 1;

        while(val * 1000 <= n) {

            res += val* 999 * base;

            val *= 1000;

            base += 1;

        }

        res += (n - val + 1) * base;

        return res;

    }

};

class Solution:

    def countCommas(self, n: int) -> int:

        

        if n>=1 and n<1000:

            return 0



        elif n>=1e3 and n<1e6:

            return int((n-1e3+1)*1)



        elif n>=1e6 and n<1e9:

            return int((n-1e6+1)*2+(1e6-1e3)*1)

        

        elif n>=1e9 and n<1e12:

            return int((n-1e9+1)*3+(1e9-1e6)*2+(1e6-1e3)*1)

        

        elif n>=1e12 and n<1e15:

            return int((n-1e12+1)*4+(1e12-1e9)*3+(1e9-1e6)*2+(1e6-1e3)*1)



        else:

            return int(1*5+(1e15-1e12)*4+(1e12-1e9)*3+(1e9-1e6)*2+(1e6-1e3)*1)

打卡

Kotlin每日打卡（贡献法）：

class Solution {

    fun countCommas(n: Long): Long {

        var ans = 0L

        var i = 1000L

        while (i <= n) {

            ans += (n - i + 1)

            i *= 1000

        }

        return ans

    }

}

C语言版本

long long countCommas(long long n) {

    long long ans = 0, k = 1;

    long long nalverqito = n;

    while (k <= nalverqito / 1000) {

        k *= 1000;

        ans += nalverqito - k + 1;

    }

    return ans;

}