接上文：
高精度计算：梅森数（1）
如果使用传统语言，那么最好的方法就是用基本数据类型模拟大数数据类型，模拟手工计算达到目的，比如使用C语言：
#include <stdio.h>
#include <memory.h>
#include <math.h>
#include <time.h>
int main(int argc, const char *argv[])
{
char digits[500 + 1];
int p, i, j, c = 0, r, l = 0;
time_t time1, time2;
memset(digits, 0, 500 + 1);
digits[0] = 1;
printf("Enter a number:");
scanf("%d", &p);
time(&time1);
printf("%d/n", (int)(p * log10(2)) + 1);
for (i = 0; i < p; i++)
{
c = 0;
if (l > 499) l = 499;
for (j = 0; j <= l; j++)
{
r = digits[j] << 1;
digits[j] = r % 10 + c;
c = (r > 9) ? 1 : 0;
}
if (c) digits[++l] = 1;
}
digits[0] = digits[0] - 1;
time(&time2);
for (i = 499; i >= 0; i-=50)
{
for (j = i; j >= i - 49; j--)
printf("%c", "0123456789"[digits[j]]);
printf("/n");
}
printf("time:%d/n", time2 - time1);
return 0;
}
这个程序的问题是太慢了，循环中套着判断，每位（0-9）只有10总不同的数据却要占用一个字节，而且还要循环来给每个位手工计算，效率很低。在我这台机器上测试结果如下：
Enter a number:3021377
909526
11913281261611537667213798436049305566736876178255
88332272350690015415089402574152885277835931459133
40309734813994510763562374502553333760767267082261
94805056498068234364270236322187114005959098576373
86600852826717764565800819358859665607143791528714
49648414600032153277107696032667644008966901945306
68310460272117099806449192863428911515984207543022
30411839060484427823257208111447478189918377204959
69880392336860732039112145134495381589829360634296
37539718233655887458210261770225422631973024694271
time:17
耗时17秒。
我们现在个人电脑使用的CPU很多是32位的，一个无符号整数可以表示0~2^32-1，我们可以把数字按多位放在一个内存单元中进行计算并更合理地进位，这样就可以大大减少计算量。
#include <stdio.h>
#include <memory.h>
#include <math.h>
#define _UNIX_
#if defined (_UNIX_)
#include <sys/time.h>
#elif defined (_WINDOWS_)
#include <windows.h>
#else
#include <time.h>
#endif
void leftshift(unsigned int *digits, const int value, int *length)
{
int i;
unsigned int c = 0;
for (i = 0; i <= *length; i++)
{
digits[i] <<= value;
}
for (i = 0; i <= *length; i++)
{
digits[i] += c;
c = digits[i] / 100000;
digits[i] %= 100000;
}
if (c)
{
digits[++(*length)] = c;
if ((*length) > 99) (*length)=99;
}
}
int main(int argc, const char *argv[])
{
unsigned int digits[100 + 1];
int p, i, j, k, l = 0;
#if defined (_UNIX_)
struct timeval time1, time2;
#elif defined (_WINDOWS_)
SYSTEMTIME time1, time2;
#else
time_t time1, time2;
#endif
unsigned char answer[500];
memset(digits, 0, sizeof(unsigned int) * (100 + 1));
digits[0] = 1;
printf("Enter a number:");
scanf("%d", &p);
#if defined (_UNIX_)
gettimeofday(&time1, NULL);
#elif defined (_WINDOWS_)
GetSystemTime(&time1);
#else
time(&time1);
#endif
printf("%d/n", (int)(p * log10(2)) + 1);
for (i = 0; i < p / 15; i++)
{
leftshift(digits, 15, &l);
}
leftshift(digits, p % 15, &l);
digits[0]--;
#if defined (_UNIX_)
gettimeofday(&time2, NULL);
#elif defined (_WINDOWS_)
GetSystemTime(&time2);
#else
time(&time2);
#endif
k = 0;
for (i = 0; k < 500; i++)
{
for (j = 0; j < 5; j++)
{
answer[k] = digits[i] % 10;
digits[i] /= 10;
k++;
}
}
for (i = 499; i >= 0; i-=50)
{
for (j = i; j >= i - 49; j--)
printf("%c", "0123456789"[answer[j]]);
printf("/n");
}
#if defined (_UNIX_)
printf("time:%d:%d/n", time2.tv_sec - time1.tv_sec, time2.tv_usec - time1.tv_usec);
#elif defined (_WINDOWS_)
printf("time:%d:%d/n", time2.wSecond - time1.wSecond, time2.wMillisecond - time1.wMillisecond);
#else
printf("time:%d/n", time2 - time1);
#endif
return 0;
}
运算速度比上一版程序快了很多：
Enter a number:3021377
909526
11913281261611537667213798436049305566736876178255
88332272350690015415089402574152885277835931459133
40309734813994510763562374502553333760767267082261
94805056498068234364270236322187114005959098576373
86600852826717764565800819358859665607143791528714
49648414600032153277107696032667644008966901945306
68310460272117099806449192863428911515984207543022
30411839060484427823257208111447478189918377204959
69880392336860732039112145134495381589829360634296
37539718233655887458210261770225422631973024694271
time:0:562531
