今天刷了高精度专题，本来准备写一个高精度类，然后直接套模版，后来发现根据每题的要求分开写高精度反而效率高。
主要涉及了高精度加法、乘法、除法、取余（其中后两项为高精度和低精度进行运算）。
终于打过一遍高精度除以低精度了，高精度专题也算圆满了。
高精度加法，水题。
#include <iostream>
#include <string>
#include <memory.h>
using namespace std;
const int MAX = 10240;
int nLen;
int pData[MAX];
int main()
{
nLen = 1;
memset(pData, 0, sizeof(pData));
string x;
while(cin >> x)
{
if(x != "0")
{
while(x[0] == ‘0‘) { x = x.substr(1, x.length() - 1); }
nLen = max(nLen, (int)x.length());
for(int i = 1; i <= x.length(); i++)
{
pData[i] += x[x.length() - i] - ‘0‘;
pData[i + 1] += pData[i] / 10;
pData[i] %= 10;
}
while(pData[nLen + 1]) { nLen++; }
}
else
{
for(int i = nLen; i >= 1; i--)
{ cout << pData[i]; }
cout << endl;
nLen = 1;
memset(pData, 0, sizeof(pData));
}
}
return 0;
}
高精度乘法，要注意的是某一个乘数为0的情况，要特判一下。
#include <iostream>
#include <string>
#include <memory.h>
using namespace std;
const int MAX = 10240;
int nLen;
int pData[MAX];
int main()
{
string x, y;
while(cin >> x >> y)
{
if(x.length() < y.length()) { swap(x, y); }
nLen = x.length() + y.length();
memset(pData, 0, sizeof(pData));
for(int i = 1; i <= y.length(); i++)
{
for(int j = 1; j <= x.length(); j++)
{
pData[i + j - 1] += (x[x.length() - j] - ‘0‘) * (y[y.length() - i] - ‘0‘);
pData[i + j] += pData[i + j - 1] / 10;
pData[i + j - 1] %= 10;
}
}
while(nLen && !pData[nLen]) { nLen--; }
if(nLen == 0) { nLen = 1; }
for(int i = nLen; i >= 1; i--)
{ cout << pData[i]; }
cout << endl;
}
return 0;
}
判断加数、乘数以及结果是否在int范围内，一开始把结果都算出来再比较。后来查了资料发现可以直接用double过，其中使用到了atof这个命令。
#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;
const int MAX_INT = 2147483647;
int main()
{
char dwOpt;
string x, y;
while(cin >> x >> dwOpt >> y)
{
cout << x << " " << dwOpt << " " << y << endl;
double a, b;
a = atof(x.c_str());
b = atof(y.c_str());
if(a > MAX_INT) { cout << "first number too big" << endl; }
if(b > MAX_INT) { cout << "second number too big" << endl; }
if(dwOpt == ‘+‘ && a + b > MAX_INT) { cout << "result too big" << endl; }
if(dwOpt == ‘*‘ && a * b > MAX_INT) { cout << "result too big" << endl; }
}
}
带有小数的乘方问题。
先确定小数点的位数，然后去掉小数点，进行高精度乘方运算，最后加上小数点即可。
#include <iostream>
#include <string>
#include <memory.h>
using namespace std;
const int MAX = 10240;
int nLen, nDot;
int pData[MAX];
string Solve(string x, string y);
int main()
{
string x;
int y;
while(cin >> x >> y)
{
while(x[x.length() - 1] == ‘0‘) { x = x.substr(0, x.length() - 1); }
nDot = x.length() - x.find(‘.‘) - 1;
nDot *= y;
x = x.substr(0, x.find(‘.‘)) + x.substr(x.find(‘.‘) + 1, x.length() - x.find(‘.‘) - 1);
string z = "1";
for(int i = 1; i <= y; i++)
{ z = Solve(z, x); }
if(z.length() <= nDot)
{
cout << ".";
for(int i = z.length(); i < nDot; i++)
{ cout << "0"; }
cout << z << endl;
}
else
{
for(int i = 0; i < z.length() - nDot; i++)
{ cout << z[i]; }
cout << ".";
for(int i = z.length() - nDot; i < z.length(); i++)
{ cout << z[i]; }
cout << endl;
}
}
return 0;
}
string Solve(string x, string y)
{
if(x.length() < y.length()) { swap(x, y); }
nLen = x.length() + y.length();
memset(pData, 0, sizeof(pData));
for(int i = 1; i <= y.length(); i++)
{
for(int j = 1; j <= x.length(); j++)
{
pData[i + j - 1] += (x[x.length() - j] - ‘0‘) * (y[y.length() - i] - ‘0‘);
pData[i + j] += pData[i + j - 1] / 10;
pData[i + j - 1] %= 10;
}
}
while(nLen && !pData[nLen]) { nLen--; }
if(nLen == 0) { nLen = 1; }
string z = "";
for(int i = 1; i <= nLen; i++)
{ z = (char)(pData[i] + ‘0‘) + z; }
return z;
}
高精度除以低精度以及取模运算。在网上搜了资料，终于学会了。但是要注意0 % X的情况。
另外想说的是非常喜欢这道题目的标题“If We Were A Child Again”，勾起了多少回忆。
#include <iostream>
#include <memory.h>
using namespace std;
const int MAX = 10240;
int nLen;
int pData[MAX], pAns[MAX];
string Solve(string x, long long y, char dwOpt);
int main()
{
char dwOpt;
string x;
long long y;
while(cin >> x >> dwOpt >> y)
{ cout << Solve(x, y, dwOpt) << endl; }
return 0;
}
string Solve(string x, long long y, char dwOpt)
{
memset(pData, 0, sizeof(pData));
memset(pAns, 0, sizeof(pAns));
nLen = x.length();
for(int i = 1; i <= x.length(); i++)
{ pData[i] = x[x.length() - i] - ‘0‘; }
long long d = 0;
for(int i = nLen; i >= 1; i--)
{
d = d * 10 + pData[i];
pAns[i] = d / y;
d %= y;
}
while(nLen && pAns[nLen] == 0) { nLen--; }
if(nLen == 0) { nLen = 1; }
string z = "";
if(dwOpt == ‘/‘)
{
for(int i = 1; i <= nLen; i++)
{ z = (char)(pAns[i] + ‘0‘) + z; }
}
else
{
while(d)
{
z = (char)(d % 10 + ‘0‘) + z;
d /= 10;
}
if(z == "") { z = "0"; }
}
return z;
}
今天由于题目相对比较少，所以做的比较快。接下来一个专题应该是排序和检索。
原文：http://www.cnblogs.com/Ivy-End/p/4279973.html