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Exponentiation（高精度）

2024-07-15 02:10:42   227人阅读

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 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

解题思路：

题目大意就是给两个数A，B，求A的B次方。唯一难点就是含有了小数点，首先得去掉小数点，但得记住小数点的位置。然后就行乘方。得出答案后再根据小数点的位置把小数点加上去。小数部分末尾的零不能留，整数部分的前导零也不能留，还有整数部分如果为0，直接输出小数点再输出小数部分。注意这几点就行了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 200;
int main()
{
    int n, ans[maxn], num, xiao[maxn], zheng[maxn];
    char R[maxn], str[10];
    while(scanf("%s", R) != EOF)
    {
        int pos = 0, k_1 = 0, k_2 = 0, k_3 = 0;
        memset(ans, 0, sizeof(ans));
        memset(xiao, 0, sizeof(xiao));
        memset(zheng, 0, sizeof(zheng));
        scanf("%d", &n);
        for(int i = 0; i < 6; i++)
        {
            if(R[i] != '.')
                str[k_1++] = R[i];
            if(R[i] == '.')
                pos = i;
        }
        str[k_1] = 0;
        sscanf(str,"%d", &num);    // 把字符串化为整数
        for(int i = 0; i < 5; i++)
            ans[i] = str[5 - i - 1] - '0';
        for(int i = 0; i < n - 1; i++)  // 求出答案
        {
            int d = 0;
            for(int j = 0; j < maxn; j++)
            {
                ans[j] = num * ans[j] + d;
                d = ans[j] / 10;
                ans[j] %= 10;
            }
        }
        bool isBegin = false;
        for(int i = 0; i < n * (5 - pos); i++)  // 找出小数部分
        {
            if(isBegin)
            {
                xiao[++k_2] = ans[i];
            }
            else if(ans[i])
            {
                xiao[0] = ans[i];
                isBegin = true;
            }
        }
        isBegin = false;
        for(int i = maxn - 1; i >= n * (5 - pos); i--)  // 找出整数部分
        {
            if(isBegin)
            {
                zheng[++k_3] = ans[i];
            }
            else if(ans[i])
            {
                zheng[0] = ans[i];
                isBegin = true;
            }
        }
        if(zheng[0])  // 判断是否要输出整数部分
        {
            for(int i = 0; i <= k_3; i++)
            printf("%d",zheng[i]);
        }
        if(xiao[k_2] == 0 && k_2 == 0) //判断是否要输出小数部分
        {
            printf("\n");
            continue;
        }
            printf(".");
            for(int i = k_2; i >= 0; i--)
                printf("%d",xiao[i]);
        printf("\n");
    }
    return 0;
}



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