1001. 求高精度幂

Description

对数值很大、精度很高的数进行高精度计算是一类十分常见的问题。比如，对国债进行计算就是属于这类问题。

现在要你解决的问题是：对一个实数R( 0.0 < R < 99.999 )，要求写程序精确计算 R 的 n 次方(Rn)，其中n 是整数并且 0 < n <= 25。

Input

T输入包括多组 R 和 n。 R 的值占第 1 到第 6 列，n 的值占第 8 和第 9 列。

Output

对于每组输入，要求输出一行，该行包含精确的 R 的 n 次方。输出需要去掉前导的 0 后不要的 0 。如果输出是整数，不要输出小数点。

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

代码

#include<iostream>#include<string>#include<vector>#include <cmath>using namespace std;char* itoa(int num, char*str, int radix){/*索引表*/    char index[] = "0123456789ABCDEF";    unsigned unum;/*中间变量*/    int i = 0, j, k;    /*确定unum的值*/    if (radix == 10 && num < 0)/*十进制负数*/    {        unum = (unsigned)-num;        str[i++] = '-';    }    else unum = (unsigned)num;/*其他情况*/                              /*转换*/    do {        str[i++] = index[unum % (unsigned)radix];        unum /= radix;    } while (unum);    str[i] = '\0';    /*逆序*/    if (str[0] == '-')k = 1;/*十进制负数*/    else k = 0;    char temp;    for (j = k; j <= (i - 1) / 2; j++)    {        temp = str[j];        str[j] = str[i - 1 + k - j];        str[i - 1 + k - j] = temp;    }    return str;}string getTimes(string s,string s2) {    string res2;    vector<string> res;    for (int i = s2.size() - 1; i >= 0; i--) {        int j = s.size() - 1, shang = 0;        string ss;        int down = s2[i] - '0';        for (; j >= 0; j--) {            int m = s[j] - '0';            int temp = m*down + shang;            if (j - 1 == -1) {                char s3[25];                ss.insert(0, itoa(temp, s3, 10));            }            else                ss.insert(ss.begin(), temp % 10 + '0');            shang = temp / 10;        }        if (res.size() > 0) {            int m = res.size();            while (m--) {                ss += '0';            }        }        //cout<<ss<<endl;                   res.push_back(ss);    }    int end = res[res.size()-1].size() - 1;    for (int m = 0; m < res.size(); m++) {        int zero = end - res[m].size() + 1;        while (zero--) {            res[m].insert(res[m].begin(), '0');        }        //cout<<res[m]<<endl;    }    int shang = 0, sum = 0;    end = res[0].size();    while (end--) {        sum = 0;        for (int m = 0; m < res.size(); m++) {            sum += (res[m][end] - '0');        }        sum += shang;        //cout << sum << endl;        if (end - 1 == -1) {            char s3[25];            res2.insert(0, itoa(sum, s3, 10));        }        else {            res2.insert(res2.begin(), sum % 10 + '0');        }        shang = sum / 10;    }    return res2;}int main() {    string s, res2;    vector<string> res;    double s4;    int n;    while (cin >> s >> n) {        int count = 0;//小数点位数         //double resl=pow(s4, n);        for (string::iterator iter = s.begin(); iter != s.end(); iter++, count++) {            if (*iter == '.') {                //cout<<"dsdsd"<<endl;                count = (s.size() - count - 1)*n;                //cout<<count;                s.erase(iter);                break;            }        }        string s1 = s;        if(n>1)            s1 = getTimes(s, s);        for (int i = 2; i < n; i++) {            s1 = getTimes(s1, s);        }        int point_pos = s1.size() - count;        s1.insert(point_pos, ".");        if (s1[0] == '0')            s1.erase(s1.begin());        while (s1[s1.size()-1] == '0') {            s1.erase(s1.end()-1);        }        if (s1[s1.size() - 1] == '.') {            s1.erase(s1.end() - 1);        }        cout << s1 << endl;    }    return 0;}