N---Exponentiation

题目描述

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 R n 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

题解：高精度问题
1.记录小数点的位置，
2.判断有无整数
3.去除无效的0；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#define maxn 10000
#define INF 0x3f3f3f3f
using namespace std;
int b[maxn + 5];
int len;
void mul(int sum){int i;int add=0;  //相乘后增加的for(i = 0; i < len; i++){int term = b[i]*sum +add;b[i] = term%10;add = term/10;}while(add){b[i++] = add%10;add = add / 10;}len = i;   //记录len的长度
}
int main()
{char a[8];int n;while(scanf("%s %d",&a,&n)!=EOF){int p = 0, sum = 0;  //p记录小数点的位置，即小数点后面有几个小数for(int i = 0; i < strlen(a); i++){if(a[i] == '.'){p = (strlen(a) - i - 1)*n;}else {sum = sum*10 + a[i] - '0';//将字符串全部转化为整数形式}}b[0] = 1;len = 1;for(int i = 0; i < n; i++)mul(sum);cout<<len<<endl;if(len <= p){  //len小于p 说明没有整数部分printf(".");for(int i = 0; i < p - len; i++){printf("0");}int k = 0;while(b[k] == 0)k++;  //去掉无效的0for(int i = len - 1; i >= k; i--)printf("%d",b[i]);printf("\n");}else {  //len>p,说明有整数，int j = 0;while(b[j] == 0 && j < p)j++;//去除无效0for(int i = len - 1; i >= j; i--){if(i + 1 == p)printf(".");printf("%d",b[i]);}printf("\n");}}return 0;
}