@Marcin Łoś 很赚钱,不使用库或滚动您自己的函数就没有 C 解决方案。
Follows 是一个有趣但没有想象力的解决方案,其中大量存储为char
(以相反顺序)的数组。
#include <stdio.h>
#include <string.h>
#include <math.h>
void Mult(char *BigNum, unsigned Factor) {
unsigned Accumulator = 0;
char Digit;
while ((Digit = *BigNum) != '\0') {
Accumulator += ((unsigned)(Digit - '0')) * Factor;
*BigNum++ = Accumulator%10 + '0';
Accumulator /= 10;
}
while (Accumulator > 0) {
*BigNum++ = Accumulator%10 + '0';
Accumulator /= 10;
}
*BigNum = '\0';
}
int main(){
unsigned N = 500;
unsigned Factor;
char BigNum[(size_t) (N*log(N) + 2)]; // Form answer, in reverse order, as a string
strcpy(BigNum, "1");
for (Factor = 1; Factor <= N; Factor++) {
Mult(BigNum, Factor);
}
printf("%u! Length:%zu Reverse:\"%s\"\n", Factor - 1, strlen(BigNum), BigNum);
unsigned long Sum = 0;
size_t i;
for (i=0; BigNum[i]; i++) {
Sum += BigNum[i] - '0';
}
printf("Sum of digits:%lu\n", Sum);
return 0;
}
500! Length:1135 Reverse:"000...221"
Sum of digits:4599