Exponentiation（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

Hint

If you don't know how to determine wheather encounted the end of input: 
s is a string and n is an integer 

C++while(cin>>s>>n){...}cwhile(scanf("%s%d",s,&n)==2) //to  see if the scanf read in as many items as you want /*while(scanf(%s%d",s,&n)!=EOF) //this also work    */{...}

解题思路

大家看AC代码就行了，再次参考了userluoxuan的，故在此放上他的链接（http://blog.csdn.net/userluoxuan/article/details/38317623）

未AC代码，样例能过，但如果次方数为1时，有bug，而且题意是必定有小数点的，看来我又理解错了，-_-!~~还是前面这篇思路更好些，

sscanf 字符串=>长整形的函数！！！！！！！

AC代码

#include <stdio.h>#include <string.h>const int maxn = 205;int main(){    int n, ans[maxn], num, xiao[maxn], zhen[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));                       //全部初始化，我总是忘了        memset(xiao, 0, sizeof(xiao));        memset(zhen, 0, sizeof(zhen));        scanf("%d", &n);        for(int i = 0; i < 6; i++)        {            if(R[i] != '.')                str[k_1++] = R[i];            if(R[i] == '.')                pos = i;                                //pos记录小数点的位置        }        str[k_1] = 0;        sscanf(str,"%d", &num);                          //字符串转换成长整形        for(int i = 0; i < 5; i++)            ans[i] = str[5 - i - 1] - '0';        for(int i = 0; i < n - 1; i++)        {            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++)        {            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--)           //注意整数和小数存储的顺序        {            if(isBegin)            {                zhen[++k_3] = ans[i];            }            else if(ans[i])            {                zhen[0] = ans[i];                isBegin = true;            }        }        if(zhen[0])        {            for(int i = 0; i <= k_3; i++)            printf("%d",zhen[i]);        }        if(xiao[k_2] == 0 && k_2 == 0)        {            printf("\n");            continue;        }        printf(".");        for(int i = k_2; i >= 0; i--)            printf("%d",xiao[i]);        printf("\n");    }    return 0;}

未AC代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 200;int main(){    char str[10];    int a[maxn], b[maxn], c[maxn], n, L, L_zhen, L_xiao, L_ans, L_a, L_b, L_c, h, i, j;    while(scanf("%s%d", str, &n) != EOF)    {        L = strlen(str);        L_xiao = 0;        for(i = L - 1, j = 0; i >= 0; i--)        {            if(str[i] == '.')                L_xiao = (5 - i) * n;            else            {                b[j] = a[j] = str[i] - '0';                j++;            }        }        L_b = L_a = j;        for(h = 2; h <= n; h++)        {            memset(c, 0, sizeof(c));            for(i = 0; i < L_a; i++)                for(j = 0; j < L_b; j++)                    c[i + j] += a[i] * b[j];            L_c = L_a + L_b - 1;            for(i = 0; i < L_c; i++)            {                c[i + 1] += c[i] / 10;                c[i] %= 10;            }            if(c[ L_c ])                L_c += 1;            for(i = 0; i < L_c; i++)                b[i] = c[i];            L_b = L_c;        }        L_zhen = L_c - L_xiao;        L_ans = L_c;        for(i = 0; i < L_c; i++)        {            if(c[i] == 0 && i + 1 <= L_xiao)                L_ans--;            if(c[i] != 0)                break;        }        for(i = L_c - 1; i >= L_c - L_ans; i--)        {            if(i == L_c - L_zhen && L_zhen == 1 && c[i] == 0)                continue;            if(i == L_c - L_zhen - 1)                printf(".%d", c[i]);            else                printf("%d", c[i]);        }        printf("\n");    }    return 0;}