UVa 748 Exponentiation解题报告

// UVa 748 Exponentiation
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;

const int maxn = 10000;
struct bign
{
int len, s[maxn];

bign()//构造函数
{
memset(s, 0, sizeof(s));
len = 1;
}

bign operator = (const char * num)//重载运算符=
{
len = strlen(num);
for (int i = 0; i < len; i++)
s[i] = num[len - i - 1] - '0';
return *this;
}

bign(const char * num){ *this = num; }//支持初始化操作

bign operator = (int num)
{
char s[maxn];
sprintf(s, "%d", num);//把num输出到字符串s中
*this = s;
return *this;
}

bign(int num){ *this = num; }

string str() const     //利用string类把字符串数组转换为字符串，方便用<<,>>输出
{
string res = "";
for (int i = 0; i < len; i++)
res = (char)(s[i] + '0') + res;
if (res == "")	res = "0";
return res;
}

bign operator + (const bign& b) const//定义加法
{
bign c;
c.len = 0;
for (int i = 0, g = 0; g || i < max(len, b.len); i++)//要两个数的位数都计算一便
{
int x = g;//g为其余数
if (i < len)	x += s[i];
if (i < b.len)	x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}

bign operator += (const bign& b)
{
*this = *this + b;
return *this;
}

void clean()//把0排除，得到真实的len
{
while (len > 1 && !s[len - 1])
len--;
}

bign operator - (const bign& b) const
{
bign c;
c.len = 0;
for (int i = 0, g = 0; i < len; i++)
{
int x = s[i] - g;//减去借1
if (i < b.len)	x -= b.s[i];//上减下
if (x >= 0)		g = 0;
else
{
g = 1;//x<0说明需要向前借1
x += 10;//将x变为正
}
c.s[c.len++] = x;
}
c.clean();//x可能为0
return c;
}

bign operator -= (const bign& b)
{
*this = *this - b;
return *this;
}

bign operator * (const bign& b) const
{
bign c;
c.len = len + b.len;//结果的位数最大为两个因子位数之和
for (int i = 0; i < len; i++)
for (int j = 0; j < b.len; j++)
c.s[i + j] += s[i] * b.s[j];//对应位置的积的和，累加起来就是结果
for (int i = 0; i < c.len - 1; i++)
{
c.s[i + 1] += c.s[i] / 10;//进位的值
c.s[i] %= 10;//余数位
}
c.clean();
return c;
}

bign operator *= (const bign& b)
{
*this = *this * b;
return *this;
}
//重载比较运算符
bool operator < (const bign& b) const
{
if (len != b.len)
return len < b.len;
for (int i = len - 1; i >= 0; i--)
if (s[i] != b.s[i])
return len < b.len;
return false;
}

bool operator >(const bign& b) const
{
return b < *this;
}

bool operator == (const bign& b) const
{
return !(b < *this) && !(b > *this);
}

bool operator != (const bign& b) const
{
return b < *this || b > *this;
}

bool operator <= (const bign& b) const
{
return !(b < *this);
}

bool operator >= (const bign& b) const
{
return !(b > *this);
}
};

//重载<<,>>，支持直接输出bign对象
istream& operator >> (istream &in, bign& x)//
{
string s;
in >> s;
x = s.c_str();
return in;
}

ostream& operator << (ostream &out, const bign& x)
{
out << x.str();
return out;
}

int main()
{
//freopen("data.txt", "r", stdin);
string a, b;
int n;
while (cin >> a >> n)
{
int i = a.find('.');//找出小数点的位置

for(int i = a.length() - 1; i >= 0; i--)//除去后面的0
if(a[i] == '0')
a.erase(i);
else
break;
b = a.substr(0, i) + a.substr(i + 1);//把小数点去掉组成新的数

const char * str = b.c_str();//转化为c字符串
bign num = str;//转化为bign对象
bign product = str;
for(int j = 1; j < n; j++)//累乘
product *= num;
int dos = (b.length() - i) * n;//得到小数点到最后非零位的长度
string c = product.str();//把bign对象转化为string对象
dos = c.length() - dos;//得到小数点的新位置
if(dos < 0)//说明没有整数部分
{
printf(".");
for(int k = dos; k < 0; k++)//把缺少的前导0补充输出
printf("0");
}

for(int j = 0; j < c.length(); j++)//输出c
if(j == dos)
{
printf(".");
j--;//找到小数点的位置后要j--，使c继续不间断输出
dos = -1;//防止无限循环，缺少这步会使条件一直成立。
}
else
cout << c[j];
cout << endl;
}
}