Exponentiation（高精度）

標籤：高精度

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

95.123 120.4321 205.1234 156.7592  998.999 101.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201

解題思路：

題目大意就是給兩個數A，B，求A的B次方。唯一痛點就是含有了小數點，首先得去掉小數點，但得記住小數點的位置。然後就行乘方。得出答案後再根據小數點的位置把小數點加上去。小數部分末尾的零不能留，整數部分的前置字元為零也不能留，還有整數部分如果為0，直接輸出小數點再輸出小數部分。注意這幾點就行了。

AC代碼：

#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);    // 把字串化為整數        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++)  // 找出小數部分        {            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--)  // 找出整數部分        {            if(isBegin)            {                zheng[++k_3] = ans[i];            }            else if(ans[i])            {                zheng[0] = ans[i];                isBegin = true;            }        }        if(zheng[0])  // 判斷是否要輸出整數部分        {            for(int i = 0; i <= k_3; i++)            printf("%d",zheng[i]);        }        if(xiao[k_2] == 0 && k_2 == 0) //判斷是否要輸出小數部分        {            printf("\n");            continue;        }            printf(".");            for(int i = k_2; i >= 0; i--)                printf("%d",xiao[i]);        printf("\n");    }    return 0;}