计算2^4000内数字0到9的分布

如题，用大数乘法很简单，但在别处看到一个用100000进制来计算的代码。如下：

#include <vector>
#include <stdio.h>
int main()
{
    std::vector<int> bigNumber(400);
    bigNumber[0] = 1;
    for (int i = 0; i < 4000; i++) {
        for (int j = 0; j < 400; j++) {
            bigNumber[j] *= 2;
        }
        for (int j = 0; j < 399; j++) {
            bigNumber[j + 1] += bigNumber[j] / 100000; // 100000进制 
            bigNumber[j] %= 100000;
        }
    }
    bool willOutput = false;
    int totalNum = 0;
    for (int j = 399; j >= 0; j--) {
        if (bigNumber[j] != 0 || willOutput) {
            totalNum += 5;
            printf(willOutput ? "%05d" : "%d", bigNumber[j]);
            willOutput = true;
        }
    }
    printf("
%d
", totalNum);
    return 0;
}