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数个数的题目脑子总是转不动

#define _CRT_SECURE_NO_WARNINGS
#include<stdlib.h>
#include<iostream>
#include<string>
#include<sstream>
#include<vector>

using namespace std;

char findchar(long long n) {
    stringstream ss;
    string str;
    char ch = 0;
    int mi = 1;
    long long seqCount = 0;

    if (n < 0) { return -1; }
    else {
        if (n == 0) {
            ch = 48;
            return ch;
        }
        else {
            while (true) {
                //0 第0 个数； 1~9，第1~9个数；10~99，第 0+9 + 1 = 10 ~9*10*2 个数；100~999第 0+9+9*10*2 + 1 =191 ~0+9+9*10*2+9*100*3 个数；
                if (seqCount + 9 * std::pow(10, mi - 1)* mi - n >= 0) {
                    break;//mi位数的共有个（0~9）数字位 9 *std::pow(10, mi-1) *mi
                }
                else {
                    seqCount += 9 * std::pow(10, mi - 1)* mi;
                    mi++;
                }
            }
        }
        //n=1，mi=1 ，seqCount=0，seqCount + 9 * std::pow(10, mi - 1)* mi = 9 > 1,跳出。
        //n=9,````````````````````````````````````````````````````````````=9 =9 跳出
        //n=10，``````````````````````````````````````````````````````````=9 不满足 >=10;
        //     mi=2, seqCount = 9`````````````````````````````````````````=189满足大于10跳出； //n=189，`````````````````````````````````````````````````````````=189满足大于10跳出;
        long long   number = std::pow(10, mi - 1) + (n - seqCount - 1) / mi;                                                                        
        long long   left = (n - seqCount - 1) % mi;
        ss << number;
        ss >> str;
        char ch = str.at(left);
        return ch;
    }
}

int main() {
        for (int i = 0; i < 192; ++i) {
            char ch = findchar(i);
            cout << "第" << i << "个数字是" << ch << endl;
        }
system("pause");
return 0;
}