PKU-ACM1001

问题描述

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.

输入

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.

输出

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

【算法设计】

  输入：以字符串和整数分别输入底和幂

  处理：以1000进制方式运算，并用额外的数组记录中间的预算结果

  输出：先以字符串方式记录最终结果，然后插入小数点，最后对字符串做裁剪处理已达到预期效果

【代码】

#include <stdio.h> #include <string.h> #include <stdlib.h> #define DEEP 50 #define ROUND(x) ((int)( (x) >= 0 ? (x) + .5 : (x) - .5)) int main(void) { char s[8],*end; //s保存输入的底 char *t; int n,i; int np,temp; int data[DEEP], data_t[DEEP], index,j,x,st; double data0; int meta; char result[200],*p,*start,*stop; while(scanf("%s %d",s,&n) != EOF){ //注意结束条件 t = strchr(s,'.'); //计算小数点后位数，存在np中 if(t) np = 5 - (t-s); else np = 0; for(i=0;i<DEEP;i++) //对data数组初始化，初始值为0 data[i] = 0; data0 = strtod(s, &end); //将字符串s转换为double值，不断乘10直到没有小数点存为meta for(i=np;i>0;i--) data0 *= 10; meta = ROUND(data0); // (1) if(meta == 0){ //若metal为0，则直接输出结果0 printf("0/n"); continue; } data [0] = meta % 1000; //以1000进制方式将meta存在data数组中，初始只在data[0]与data[1]中 data[1] = meta / 1000; index = 1; //index记录data数组中最高有效数的下标 for(i=1;i<n;i++){ //循环乘meta值(n-1)次 for(j=0;j<DEEP;j++) //每次循环计算前都要对data_t数组清0 data_t[j] = 0; for(j=0;j<=index;j++){ //以data_t数组保存每次计算结果 temp = data[j] * meta; x=j; while(temp){ st = data_t[x] + temp % 1000; // (2) data_t[x] = st % 1000; data_t[x+1] = data_t[x+1] + st / 1000; temp = temp / 1000; x++; } } x = x+1; //修正index值，使其至少为指向最高有效数的小标 if(x > index) index = x; for(j=0;j<=index;j++) //将结果从data_t数组导到data数组 data[j] = data_t[j]; } p = result; //利用sprintf函数将最终结果以字符串的形式存在result数组中 for(i=0;i<n*np;i++,p++) // (3) *p = '0'; for(i=index;i>=0;i--){ sprintf(p,"%03d",data[i]); p = p+3; } *p = '/0'; stop = p; for(i=0;i<=n*np;i++){ //根据n与np值，将小数点插在result数组的合适位置 *(p+1) = *p; p--; } *(p+1) = '.'; start = result; //将result数组做修剪操作，剪去前后的'0' while(*start == '0') start++; while(*stop == '0') *stop-- = '/0'; if(*stop == '.') // (4) *stop = '/0'; printf("%s/n",start); } // end of while return 0; } 

（1）由于实数在计算机上的表示不确定，从实数向整数转换时不能直接做类型转换；可以使用上面类似的四舍五入法

（2）运算过程中，计算中间结果时注意进位

（3）由于不能确定底值（比如，0.0001），我们在数串前加额外的'0'

（4）不要忘记最终结果中小数点可能在最后（比如，112341.）的情况