该程序应该给出任何大小的阶乘的最后 100 位数字。然而，main() 中的 counter2++ 发生了一些奇怪的事情。每次循环在 main() 函数中运行（即 99 次），counter2 都会增加 +1。但是，这是显示的内容：
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
71
86
90
123
164
196
207
254
300
362
432
471
551
620
630
708
761
772
857
896
985
1036
1100
1116
1207
1209
1280
1356
1417
1452
1512
Counter2 最终是 1512 而不是 100，但是如果我从 main() 中删除 mult(i) 或 carry()，那么它会显示 100。为什么 counter2 最终是 1512 而不是 100？
#include <iostream>
using namespace std;
int numbers[100];
int counter2 = 0;
void init(){
//sets elements 1-99 of numbers[] to 0, increments counter2 by 1, sets numbers[0] = 1
for (int i = 1; i < 100; i++){
numbers[i] = 0;
}
numbers[0] = 1;
counter2++;
}
void mult(int x){
//multiplies each element by 1 through n to calculate for !n
//this is used to represent a very large number without using a BigInt library
//the nth element is a placeholder for the n+1 position of the number
//e.g 2nd element represents 100-900 of the number, 4th represents 1000-9000, etc
//carry() is used to take care of overflow, so that it's only 1 digit per element
for (int i = 0; i < 100; i++){
numbers[i] *= x;
}
}
void carry(){
//in order to make previous function work, this adds any overflow to the next
//element. e.g: 8 * 4 = 32, 3 is added to numbers[i+1], sets numbers[i] to 2
int counter = 0;
for (int i = 0; i < 100; i++){
if (numbers[i] >= 10){
counter = numbers[i] / 10;
numbers[i+1] += counter;
numbers[i] = numbers[i] % (counter * 10);
}
}
}
int main()
{
init();
for (int i = 2; i < 101; i++){
//calculates the last 100 digits of !100, but counter2 ends up being 1512
mult(i);
carry();
counter2++;
cout << counter2 << endl;
}
}