fre*_*low 14
double完全能够存储多达1023的两个大国的究竟.不要让别人告诉你浮点数在某种程度上总是不准确.这是一个特殊情况,他们不是!
double x = 1.0;
for (int n = 0; n <= 200; ++n)
{
printf("2^%d = %.0f\n", n, x);
x *= 2.0;
}
该程序的一些输出:
2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
...
2^196 = 100433627766186892221372630771322662657637687111424552206336
2^197 = 200867255532373784442745261542645325315275374222849104412672
2^198 = 401734511064747568885490523085290650630550748445698208825344
2^199 = 803469022129495137770981046170581301261101496891396417650688
2^200 = 1606938044258990275541962092341162602522202993782792835301376
只需等待256位编译器,然后使用int:-)
不,严重的是,因为你只想从1开始并保持加倍,你最好的选择是获得一个像GNU MP这样的大整数库.
你会用一段代码(未经测试)来做到这一点:
#include <stdio.h>
#include "gmp.h"
int main (void) {
int i;
mpz_t num;
mpz_init_set_ui (num, 1);
for (i = 0; i <= 200; i++) {
printf ("2^%d = ", i);
mpz_out_str (NULL, 10, num);
printf ("\n");
mpz_mul_ui (num, num, 2);
}
return 0;
}
您只需要两个操作,双重和打印就可以编写自己的long数组结构,但我认为使用GMP要容易得多.
如果你也想推出自己的,看看这个.这是我过去开发的一些大型整数库的变化/简化:
#include <stdio.h>
#include <stdlib.h>
// Use 16-bit integer for maximum portability. You could adjust
// these values for larger (or smaller) data types. SZ is the
// number of segments in a number, ROLLOVER is the maximum
// value of a segment plus one (need to be less than the
// maximum value of your datatype divided by two. WIDTH is
// the width for printing (number of "0" characters in
// ROLLOVER).
#define SZ 20
#define ROLLOVER 10000
#define WIDTH 4
typedef struct {
int data[SZ];
} tNum;
// Create a number based on an integer. It allocates the segments
// then initialises all to zero except the last - that one is
// set to the passed-in integer.
static tNum *tNumCreate (int val) {
int i;
tNum *num = malloc (sizeof (tNum));
if (num == NULL) {
printf ("MEMORY ERROR\n");
exit (1);
}
for (i = 0; i < SZ - 1; i++) {
num->data[i] = 0;
}
num->data[SZ-1] = val;
}
// Destroy the number. Simple free operation.
static void tNumDestroy (tNum *num) {
free (num);
}
// Print the number. Ignores segments until the first non-zero
// one then prints it normally. All following segments are
// padded with zeros on the left to ensure number is correct.
// If no segments were printed, the number is zero so we just
// output "0". Then, no matter what, we output newline.
static void tNumPrint (tNum *num) {
int i, first;
for (first = 1, i = 0; i < SZ; i++) {
if (first) {
if (num->data[i] != 0) {
printf ("%d", num->data[i]);
first = 0;
}
} else {
printf ("%0*d", WIDTH, num->data[i]);
}
}
if (first) {
printf ("0");
}
printf ("\n");
}
// Double a number. Simplified form of add with carry. Carry is
// initialised to zero then we work with the segments from right
// to left. We double each one and add the current carry. If
// there's overflow, we adjust for it and set carry to 1, else
// carry is set to 0. If there's carry at the end, then we have
// arithmetic overflow.
static void tNumDouble (tNum *num) {
int i, carry;
for (carry = 0, i = SZ - 1; i >= 0; i--) {
num->data[i] = num->data[i] * 2 + carry;
if (num->data[i] >= ROLLOVER) {
num->data[i] -= ROLLOVER;
carry = 1;
} else {
carry = 0;
}
}
if (carry == 1) {
printf ("OVERFLOW ERROR\n");
exit (1);
}
}
// Test program to output all powers of 2^n where n is in
// the range 0 to 200 inclusive.
int main (void) {
int i;
tNum *num = tNumCreate (1);
printf ("2^ 0 = ");
tNumPrint (num);
for (i = 1; i <= 200; i++) {
tNumDouble (num);
printf ("2^%3d = ", i);
tNumPrint (num);
}
tNumDestroy (num);
return 0;
}
及其相关输出:
2^ 0 = 1
2^ 1 = 2
2^ 2 = 4
2^ 3 = 8
2^ 4 = 16
2^ 5 = 32
2^ 6 = 64
2^ 7 = 128
2^ 8 = 256
2^ 9 = 512
: : : : :
2^191 = 3138550867693340381917894711603833208051177722232017256448
2^192 = 6277101735386680763835789423207666416102355444464034512896
2^193 = 12554203470773361527671578846415332832204710888928069025792
2^194 = 25108406941546723055343157692830665664409421777856138051584
2^195 = 50216813883093446110686315385661331328818843555712276103168
2^196 = 100433627766186892221372630771322662657637687111424552206336
2^197 = 200867255532373784442745261542645325315275374222849104412672
2^198 = 401734511064747568885490523085290650630550748445698208825344
2^199 = 803469022129495137770981046170581301261101496891396417650688
2^200 = 1606938044258990275541962092341162602522202993782792835301376
python支持开箱即用的大整数.在任何linux提示符下,运行以下命令:
$ python -c "for power in range(201): print power, 2**power"
0 1
1 2
2 4
3 8
4 16
5 32
6 64
<snip>
196 100433627766186892221372630771322662657637687111424552206336
197 200867255532373784442745261542645325315275374222849104412672
198 401734511064747568885490523085290650630550748445698208825344
199 803469022129495137770981046170581301261101496891396417650688
200 1606938044258990275541962092341162602522202993782792835301376
如有必要,可以很容易地将其制作成脚本.看任何python教程.
使用java.math.BigInteger.shiftLeft.
for (int i = 0; i <= 200; i++) {
System.out.format("%d = %s%n", i, BigInteger.ONE.shiftLeft(i));
}
输出摘录:
0 = 1
1 = 2
2 = 4
3 = 8
4 = 16
:
197 = 200867255532373784442745261542645325315275374222849104412672
198 = 401734511064747568885490523085290650630550748445698208825344
199 = 803469022129495137770981046170581301261101496891396417650688
200 = 1606938044258990275541962092341162602522202993782792835301376
如果BigInteger不可用,您也可以手动执行乘法并将其存储在a中String.
String s = "1";
for (int i = 0; i < 200; i++) {
StringBuilder sb = new StringBuilder();
int carry = 0;
for (char ch : s.toCharArray()) {
int d = Character.digit(ch, 10) * 2 + carry;
sb.append(d % 10);
carry = d / 10;
}
if (carry != 0) sb.append(carry);
s = sb.toString();
System.out.format("%d = %s%n", i + 1, sb.reverse());
}
(见完整输出)