杭电 1063 Exponentiation

Exponentiation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5265 Accepted Submission(s): 1430

Problem Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

Sample Input

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

Source

East Central North America 1988

Recommend

PrincetonBoy

这题有点难搞，主要是一个大整数的连乘，刚开始写了一个大整数相乘的函数，发现不对啊！然后就改，调试了挺长时间的，最终AC！觉得有点幸运~~~

View Code

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int num[10], result[200], len, flag = 0, cnt1;
 6 
 7 int char_to_int( char *s )
 8 {
 9     int i, dian = 0, j, jihao;
10     memset(num,0,10*sizeof(int));
11     len = strlen(s);
12     //printf( "%s\n", s );
13     //printf( "%d\n", len );
14     cnt1 = 0;
15     for( j = len - 1; j >= 0; j-- )
16          if( s[j] == '.' )
17              jihao = j;
18 
19         for( j = len - 1; j > jihao; j-- )
20              if( s[j] != '0' )
21                  break;
22         //printf( "%d\n", j )
23         
24     
25     for( i = j; i >= 0; i--)
26          if( s[i] != '.' )
27              num[cnt1++] = s[i] - '0';
28          else
29              dian = j  - i;
30     /*for( i = 0; i < cnt1; i++ )
31          printf( "%d", num[i] );
32     printf( "\n%d\n", dian );*/
33     return dian;
34 }
35 int bignummulti(  )
36 {
37     int i, j, temp[200];
38     memset(temp, 0 ,200*sizeof(int));
39     for( i = 0; i < len; i++ )
40          for( j = 0; j < 200; j++ )
41              temp[i+j] += result[j] * num[i];
42     
43     for( i = 0; i < 200; i++ )
44          if( temp[i] >= 10 )
45          {
46              temp[i+1] += temp[i] / 10;
47              temp[i] = temp[i] % 10;
48          }
49     for( i = 0; i < 200; i++ )
50         result[i] =  temp[i];
51     return 0;
52 }
53 
54 int main(int argc, char *argv[])
55 {
56     int n, dian, i, cnt, j;
57     char s[10];
58     memset(s,0,sizeof(s));
59     while( scanf( "%s %d", &s, &n )!=EOF )
60     {
61            dian = char_to_int( s ) * n;
62            memset(result,0,200 * sizeof(int));
63            result[0] = 1;
64            for( i = 1; i <= n; i++  )
65                 bignummulti();
66            
67           for( i = 199; i >= 0; i-- )
68                 if( result[i] != 0 )
69                     break;
70            for( j = 0; j < dian; j++ )
71                 if( result[j] != 0 )
72                     break;
73            if( dian > i )
74            {
75                if( i == -1 )
76                    printf( "0" );
77                else
78                    printf( "." );
79                cnt = dian - (i + 1);
80                while( cnt-- )
81                       printf( "0" );
82                for( ; i >= j; i-- )
83                     printf( "%d", result[i] );
84            }
85            else
86                for( ; i >= j; i-- )
87                {
88                     if( i == dian-1 )
89                         printf( "." );
90                     printf( "%d", result[i] );
91                }
92            printf( "\n" );     
93     }
94   
95   //system("PAUSE");    
96   return 0;
97 }