[POJ] #1001# Exponentiation : 大数乘法

一. 题目

Exponentiation

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 156373		Accepted: 38086

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

Hint

If you don't know how to determine wheather encounted the end of input:
s is a string and n is an integer

C++

while(cin>>s>>n)

{

...

}

c

while(scanf("%s%d",s,&n)==2) //to  see if the scanf read in as many items as you want

/*while(scanf(%s%d",s,&n)!=EOF) //this also work    */

{

...

}

Source

East Central North America 1988

 

二. 题意

• 给出字符串表示的浮点数 和 n次方数
• 输出其n次方运算后的值

三. 分析

• 算法核心:
  • 大数乘法：利用程序模拟算数运算过程，算出乘积
• 实现细节:
  • 将输入数字从低位到高位反转存储，计算时方便进位
  • 模拟计算时，去除小数点，统一按照大整数相乘
  • 输出时逆序输出，同时计算小数点位置，相应位输出小数点

 

三. 题解

 #include <stdio.h>
 #include <memory.h>

 #define MAXN 100
 int a[MAXN], b[MAXN], len_a, len_b;

 /* 模拟算数运算过程，算出乘积 */
 void multiply(int *a, int *b)
 {
     int i, j, len, c[MAXN];

     memset(c, , sizeof(c));

     for (i = ; a[i] != -; i++)
         for (j = ; b[j] != -; j++)
                 c[i + j] += a[i] * b[j];

     len = i + j - ;

     for (i = ; i < len; i++)
         if (c[i] >= ) { c[i + ] += c[i] / ; c[i] = c[i] % ;}

     if (c[len] != ) len_b = len + ;
     else len_b = len;

     for (i = ; i < len_b; i++) b[i] = c[i];
 }

 void main()
 {
     char str[MAXN];
     int i, j, exp;
     int st, pt, ed;

     freopen( "input.txt", "r" , stdin);

     while (scanf("%s %d", str, &exp) != EOF) {
         memset(a, -, sizeof(a));
         memset(b, -, sizeof(b));
         len_a = ;
         st = pt = ed = -;
 　　　
 　　   /*
 　　　　1.st为输入数字的启始位置
 　　　　2.pt为输入数字小数点的位置
 　　　　3.ed为输入数字的结束位置
 　　　　4.此三个中间变量的作用：
 　　　　　　4.1 去掉开头和结尾的 0
 　　　　　　4.2 对于小数点的特殊处理
 　　　　　　4.3 用于计算出乘积后小数点的正确位置
         */
         for (i = ; str[i] != '\0'; i++) {
             len_a++;

             if (str[i] != '' && str[i] != '.' && st == -) st = i;
             if (str[i] != '' && str[i] != '.' || (str[i] != '.' && pt == -)) ed = i;
             if (str[i] == '.') pt = i;
         }

         for (i = ed, j = ; i >= st; i--) {
             if (str[i] == '.') continue;

             b[j] = a[j] = str[i] - '';
             j++;
         }

         len_b = len_a = j;
         for(i = ; i < exp; i++) multiply(a, b);

         if (pt < st) {
             printf(".");
             for (i = ; i < (ed - pt) * exp - len_b; i++) printf("");
         }

         for (i = len_b - ; i >= ; i--) {
             if (pt > st && pt < ed && i == (ed - pt) * exp - ) printf(".");
             printf("%d", b[i]);
         }

         printf("\n");
     }
  }