POJ 1001 Exponentiation

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

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

这是我做POJ的第一道题，看到Accept十分兴奋！终于品尝到AC的快乐！

#include<iostream>#include<cstring>using namespace std;//用结构体来使代码更具可读性和可重复利用性structBigfloat{intpoint;//记录小数点后的位数intlen;//记录长度（除去小数点）intdata[1000];//储存数据的数组//格式化输出函数void print(){intflag = 0;while(data[flag]==0&&flag < point)    //设置标记点除去小数点后一些没用的0{      //同时当该数位整数时，也会自动去掉小数点++flag;}if(data[len - 1] == 0)--len;  //除去首位为0for(int i = len - 1;i >= flag;--i){if(i == point - 1)cout << ".";cout << data[i];}cout << endl;}};//高精度小数乘法的模板BigfloatMultiply(Bigfloat a , Bigfloat b){Bigfloattemp;temp.len = a.len + b.len;temp.point = a.point + b.point;memset(temp.data , 0 , sizeof(temp.data));for(int i = 0;i < a.len;i++)for(int j = 0;j < b.len;j++){temp.data[i + j] += a.data[i] * b.data[j];temp.data[i + j + 1] += temp.data[i + j] / 10;temp.data[i + j] %= 10;}if(temp.data[temp.len - 1] == 0) temp.len --;return temp;}//倒位的函数，方便调用BigfloatChange(char s[]){Bigfloatdata;data.len = strlen(s) - 1;memset(data.data, 0 , sizeof(data.data));for(int i = 0,j = 0;i < strlen(s);++i,++j){if(s[strlen(s) - i - 1] == '.'){++i;data.data[j] = s[strlen(s) - i - 1] - '0';data.point = j;}else{data.data[j] = s[strlen(s) - i - 1] - '0';}}return data;}intmain(){intn;Bigfloata,b;charnum[1000];while(cin >> num >> n)//通过循环利用模板完成幂乘{a = Change(num);b = a;for(int i = 1;i < n;++i){Multiply(a,b);a = Multiply(a,b);}a.print();}return 0;}