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;
}