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. 题不难。就是麻烦。。 先忽略小数点计算一个值，最后再把小数点加进去、 注意的地方： 1.输出格式； 2.忽略小数上没用的0； 3.不能忽略小数里开头的0，比如说.0023中的00； 4.整数不需要输出小数点； 5.注意输入为0的时候。 程序： /*poj 1001 exponentiation 2011-2-17*/ #include <stdio.h> #include <math.h> #include <string.h> #define true 1 #define false 0 int n,decimal_num,integer_num,raw_num,max; int result[1000]; char point=true; char raw_data[10]; void multiplication(int a){ int i,carry=0; for(i=1;i<=max;i++){ result[i]= result[i]*a+carry; carry=result[i]/10; result[i]%=10; } while(carry!=0){ max++; result[max]=carry; carry/=10; } } int main(){ int i,j; while(scanf("%s%d",raw_data,&n)==2){ raw_num=0; j=0; point=true; /*point为判断是否有小数点*/ for(i=5;i>=0;i--){ if(raw_data[i]== '.'){ decimal_num= 5-i; /*记录其小数个数*/ continue; } raw_num= raw_num+ (raw_data[i]-'0')*ceil(pow(10,j)); /*raw_num为输入去掉小数点后数值，这样写高精和整型的乘法，个人感觉简单点*/ j++; } decimal_num *=n; /*计算结果的小数个数*/ result[1]=1; max=1; /*结果一共有几个数字*/ for(i=1;i<=n;i++) multiplication(raw_num); i=1; integer_num= max-decimal_num; /*计算整数个数*/ while(result[i]==0){ /*去掉末尾0，这里需要注意，如果结果为10.00，那么整数部分的0不能去，同时，最后输出也不要输出小数点*/ if(i>decimal_num){ point=false; /*point表示是否输出小数点*/ break; } i++; } if(raw_num==0){ /*输入0实在不知道怎么处理。。。直接特判一下。。。*/ printf("0/n"); continue; } if(decimal_num>=max){ /*如果结果为.00023，需要输出小数点后必要的0*/ printf("."); point=false; for(;decimal_num!=max;decimal_num--) printf("0"); } for(;max>=i;max--){ /*输出结果*/ printf("%d",result[max]); integer_num--; if(integer_num==0 && point) printf("."); } printf("/n"); } return 0; }