作者：Leon916 来源：C++博客 酷勤网收集 2008-05-23
摘要
对数值很大、精度很高的数进行高精度计算是一类十分常见的问题。比如，对国债进行计算就是属于这类问题。现在要你解决的问题是：对一个实数R( 0.0 < R < 99.999 )，要求写程序精确计算 R 的 n 次方(Rn)，其中n 是整数并且 0 < n
求高精度幂
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 33587 | Accepted: 7590 |
Description
对数值很大、精度很高的数进行高精度计算是一类十分常见的问题。比如，对国债进行计算就是属于这类问题。
现在要你解决的问题是：对一个实数R( 0.0 < R < 99.999 )，要求写程序精确计算 R 的 n 次方(Rn)，其中n 是整数并且 0 < n <= 25。
现在要你解决的问题是：对一个实数R( 0.0 < R < 99.999 )，要求写程序精确计算 R 的 n 次方(Rn)，其中n 是整数并且 0 < n <= 25。
Input
T输入包括多组 R 和 n。 R 的值占第 1 到第 6 列，n 的值占第 8 和第 9 列。
Output
对于每组输入，要求输出一行，该行包含精确的 R 的 n 次方。输出需要去掉前导的 0 后不要的 0 。如果输出是整数，不要输出小数点。
Sample Input
95.123 12 0.4321 20 5.1234 15 6.7592 9 98.999 10 1.0100 12
Sample Output
548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201
Source
Translator
北京大学程序设计实习，Xie Di
这道题我做的很郁闷，代码也写的不好、很长，方法也一般，有点麻烦，自己都觉得很烂，在提交的时候，发现别人的代码写的很简练，但是不知道自己该怎么改进，希望大家多多提意见，谢谢大家了！
1
#include <string.h>
2#include <stdio.h>
3#define LENGTH 100
4
5typedef struct Number
6
{
7
int number;
8 int point;
9 int power;
10
}Number;
11
12void ConverTo(char *ch, int *real, int *p)
13{
14 int b = 1 ,sign = 1;
15 char *c = ch;
16 *real = 0;
17 *p = 0;
18 for( ; *c != '\0' && *c != '.'; c++)
19 *real = (*real)*10 + (*c - '0');
20 if( *c == '\0')
21
{
22 *p = 0;
23
}
24 else
25 {
26 *p = c-ch;
27 while( *(++c) != '\0')
28 *real = (*real)*10 + (*c - '0');
29 *p = c-ch-1- *p;
30 }
31}
32
33void Caculate(int *number, int real, int *p, int n, int size)
34{
35 int sum = 0;
36 int i ,t1, t2;
37 t1 = t2 = 0;
38 while( n-- > 0)
39 {
40 for(i = size-1; i >= 0; i--)
41 {
42 t1 = number[i]*real +t2;
43 number[i] = t1 % 10000;
44 t2 = t1 / 10000;
45 }
46 sum += *p;
47 }
48 *p = sum;
49}
50
51void Convert(char *ch, int *index, int *number, int *left, int right)
52{
53 int temp;
54
55 while(*left <= right)
56 {
57 temp = number[(*left)++];
58 ch[(*index)++] = temp/1000 + '0';
59 temp -= (temp/1000) * 1000;
60 ch[(*index)++] = temp/100 + '0';
61 temp -= (temp/100) * 100;
62 ch[(*index)++] = temp/10 + '0';
63 ch[(*index)++] = temp%10 + '0';
64
65 }
66}
67void ConverToChar(char *ch, int length,int *number, int left, int right, int point)
68{
69 int last,index, temp, p;
70 index = 0;
71 p = point/4;
72
73 Convert(ch, &index, number, &left, p-1);
74 temp = number[left++];
75 for(last = 0,p = 1000; last < 4; last++, p /= 10)
76 {
77 if( (4*(left-1) + last) == point)
78 ch[index++] = '.';
79 ch[index++] = temp/p + '0';
80 temp -= (temp/p) * p;
81 }
82 Convert(ch, &index, number, &left, right);
83}
84
85void Print(int *number, int p, int size)
86{
87 int left, right,index;
88 char ch[200];
89
90 index = (size*4-p)/4;
91 memset(ch, 0, sizeof(ch));
92
93 for(left = 0; left < index; left++)
94 {
95 if(number[left] == 0)
96 continue;
97 else
98 break;
99 }
100 for(right = size-1; right > index; right--)
101 {
102 if(number[right] == 0)
103 continue;
104 else
105 break;
106 }
107 ConverToChar(ch, sizeof(ch)-1, number, left, right, size*4-p);
108 for(left = 0; ch[left] == '0'; left++)
109 ;
110 for(right = sizeof(ch)-1; ch[right] == '\0' || ch[right] == '0'; right--)
111 ;
112 if(ch[right] == '.')
113 right--;
114 while(left <= right)
115 printf("%c", ch[left++]);
116 printf("\n");
117}
118
119int main(int argc, char* argv[])
120{
121
122 char ch[7];
123 Number n[LENGTH];
124 int i = 0, count = 0, p = 0;
125 int number[30];
126 memset(ch, 0, sizeof(ch));
127 while(scanf("%s %d", ch, &p) == 2)
128 {
129 ConverTo(ch, &n[count].number, &n[count].point);
130 n[count].power = p;
131 if(!(p<= 25 && p >0) || !(n[count].number> 0 && n[count].number < 99999))
132 {
133 break;
134 }
135 memset(ch,0, sizeof(ch));
136 count++;
137 }
138 while(i < count)
139 {
140 memset(number, 0, sizeof(number));
141 number[sizeof(number)/sizeof(int)-1] = 1;
142 Caculate(number, n[i].number, &n[i].point, n[i].power, sizeof(number)/sizeof(int));
143 Print(number,n[i].point,sizeof(number)/sizeof(int));
144 i++;
145 }
146 return 1;
147}
#include <string.h>2
#include <stdio.h>3
#define LENGTH 1004
5
typedef struct Number6
{7
int number;8
int point;9
int power;10
}Number;11
12
void ConverTo(char *ch, int *real, int *p)13
{14
int b = 1 ,sign = 1;15
char *c = ch;16
*real = 0;17
*p = 0;18
for( ; *c != '\0' && *c != '.'; c++)19
*real = (*real)*10 + (*c - '0');20
if( *c == '\0')21
{22
*p = 0;23
}24
else25
{26
*p = c-ch;27
while( *(++c) != '\0')28
*real = (*real)*10 + (*c - '0');29
*p = c-ch-1- *p;30
}31
}32
33
void Caculate(int *number, int real, int *p, int n, int size)34
{35
int sum = 0;36
int i ,t1, t2;37
t1 = t2 = 0;38
while( n-- > 0)39
{40
for(i = size-1; i >= 0; i--)41
{42
t1 = number[i]*real +t2;43
number[i] = t1 % 10000;44
t2 = t1 / 10000;45
}46
sum += *p;47
}48
*p = sum;49
}50
51
void Convert(char *ch, int *index, int *number, int *left, int right)52
{53
int temp;54
55
while(*left <= right)56
{57
temp = number[(*left)++];58
ch[(*index)++] = temp/1000 + '0';59
temp -= (temp/1000) * 1000;60
ch[(*index)++] = temp/100 + '0';61
temp -= (temp/100) * 100;62
ch[(*index)++] = temp/10 + '0';63
ch[(*index)++] = temp%10 + '0';64
65
}66
}67
void ConverToChar(char *ch, int length,int *number, int left, int right, int point)68
{69
int last,index, temp, p;70
index = 0;71
p = point/4;72
73
Convert(ch, &index, number, &left, p-1);74
temp = number[left++];75
for(last = 0,p = 1000; last < 4; last++, p /= 10)76
{77
if( (4*(left-1) + last) == point)78
ch[index++] = '.';79
ch[index++] = temp/p + '0';80
temp -= (temp/p) * p;81
}82
Convert(ch, &index, number, &left, right);83
}84
85
void Print(int *number, int p, int size)86
{87
int left, right,index;88
char ch[200];89
90
index = (size*4-p)/4;91
memset(ch, 0, sizeof(ch));92
93
for(left = 0; left < index; left++)94
{95
if(number[left] == 0)96
continue;97
else98
break;99
}100
for(right = size-1; right > index; right--)101
{102
if(number[right] == 0)103
continue;104
else105
break;106
}107
ConverToChar(ch, sizeof(ch)-1, number, left, right, size*4-p);108
for(left = 0; ch[left] == '0'; left++)109
;110
for(right = sizeof(ch)-1; ch[right] == '\0' || ch[right] == '0'; right--)111
;112
if(ch[right] == '.')113
right--;114
while(left <= right)115
printf("%c", ch[left++]);116
printf("\n");117
}118
119
int main(int argc, char* argv[])120
{121
122
char ch[7];123
Number n[LENGTH];124
int i = 0, count = 0, p = 0;125
int number[30];126
memset(ch, 0, sizeof(ch));127
while(scanf("%s %d", ch, &p) == 2)128
{129
ConverTo(ch, &n[count].number, &n[count].point);130
n[count].power = p;131
if(!(p<= 25 && p >0) || !(n[count].number> 0 && n[count].number < 99999))132
{133
break;134
}135
memset(ch,0, sizeof(ch));136
count++;137
}138
while(i < count)139
{140
memset(number, 0, sizeof(number));141
number[sizeof(number)/sizeof(int)-1] = 1;142
Caculate(number, n[i].number, &n[i].point, n[i].power, sizeof(number)/sizeof(int)); 143
Print(number,n[i].point,sizeof(number)/sizeof(int));144
i++;145
}146
return 1;147
}来自：1001题探讨