zoukankan      html  css  js  c++  java
  • 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
    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
     
    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 }
  • 相关阅读:
    从原理上理解NodeJS的适用场景
    core 基本操作
    SQL Server 触发器
    Centos 7 Apache .netcore 做转发
    Windows Server 使用 WAMP 并配置 Https
    centos7 apache php git pull
    Visual StudioTools for Unity 使用技巧2
    如何实现Windows Phone代码与Unity相互通信(直接调用)
    关于NGUI与原生2D混用相互遮盖的问题心得
    关于NGUI制作图集在低内存设备上的注意事项
  • 原文地址:https://www.cnblogs.com/yizhanhaha/p/3070802.html
Copyright © 2011-2022 走看看