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：

 1 #include <vector>
 2 #include <stdio.h>
 3 int main()
 4 {
 5     std::vector<int> bigNumber(400);
 6     bigNumber[0] = 1;
 7     for (int i = 0; i < 4000; i++) {
 8         for (int j = 0; j < 400; j++) {
 9             bigNumber[j] *= 2;
10         }
11         for (int j = 0; j < 399; j++) {
12             bigNumber[j + 1] += bigNumber[j] / 100000; // 100000进制
13             bigNumber[j] %= 100000;
14         }
15     }
16     bool willOutput = false;
17     int totalNum = 0;
18     for (int j = 399; j >= 0; j--) {
19         if (bigNumber[j] != 0 || willOutput) {
20             totalNum += 5;
21             printf(willOutput ? "%05d" : "%d", bigNumber[j]);
22             willOutput = true;
23         }
24     }
25     printf("\n%d\n", totalNum);
26     return 0;
27 }