高精度 - SGU 112 a^b-b^a

Problem's Link

[b]Mean:[/b]

略

[b]analyse:[/b]

简单题，只用编个高精度乘法和减法即可.

[b]Time complexity: O(N)[/b]

[b]view code[/b]

java

import java.math.BigInteger;
import java.util.Scanner;
public class Solution
{
public static void main(String[] args)
{
Scanner cin = new Scanner(System.in);
int a, b;
while(cin.hasNext())
{
a = cin.nextInt();
b = cin.nextInt();
System.out.println(BigInteger.valueOf(a).pow(b).add(BigInteger.valueOf(b).pow(a).negate()));
}
}
}

C++

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-01-06-20.55
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);

#define inf 0xfffffff
#define CLR(a,b) memset((a),(b),sizeof((a)))

using namespace std;
int const nMax = 1000;
int const base = 10;
typedef int LL;
typedef pair<LL,LL> pij;

struct Int
{
int a[nMax];
int len;
void clear()
{
CLR(a,0);
len=0;
return ;
}
Int (int n=0)
{
clear();
while(n)
{
a[len++]=n%base;
n/=base;
}
}
bool operator < (const Int& b)const
{
if(len<b.len)return true;
if(len>b.len)return false;
for(int i=len-1; i>=0; i--)
{
if(a[i]<b.a[i])return true;
if(a[i]>b.a[i])return false;
}
return false;
}
Int operator +(Int&b)
{
Int c;
c.len=max(len,b.len);
int d=0;
for(int i=0; i<c.len; i++)
{
c.a[i]=(a[i]+b.a[i]+d)%base;
d=(a[i]+b.a[i]+d)/base;
}
while(d)
{
c.a[c.len++]=d%base;
d/=base;
}
return c;
}
Int operator -(Int&b)
{
Int c;
c.len=len;
int d=0;
int f;
for(int i=0; i<c.len; i++)
{
c.a[i]=a[i]-b.a[i]-d;
d=0;
if(c.a[i]<0)
{
c.a[i]+=base;
d++;
}
// printf("c[%d]=%d\n",i,c.a[i]);
}
while(c.a[c.len-1]==0&&c.len>0)
{
c.len--;
}
if(c.len==0)c.len=1;
return c;
}
Int operator *(int &b)
{
Int c;
int d;
c.len=len;
for(int i=0; i<c.len; i++)
{
c.a[i]=(a[i]*b+d)%base;
d=(a[i]*b+d)/base;
}
while(d)
{
c.a[c.len++]=d%base;
d/=base;
}
return c;
}
Int operator *(Int & b)
{
Int c;
for(int i=0; i<len; i++)
{
for(int j=0; j<b.len; j++)
{
c.a[i+j]+=a[i]*b.a[j];
}
}
c.len=len+b.len-1;
int d=0; //就是这里忘了置0
int f;
for(int i=0; i<c.len; i++)
{
f=d+c.a[i];
c.a[i]=f%base;
d=f/base;
}
while(d)
{
c.a[c.len++]=d%base;
d/=base;
}
return c;
}
void output()
{
for(int i=len-1; i>=0; i--)printf("%d",a[i]);
printf("\n");
return ;
}
string Int2Str()
{
string ans;
ans.clear();
for(int i=len-1; i>=0; i--)
{
ans+=('0'+a[i]);
}
return ans;
}
};

Int exp(Int b,int n)
{
if(n==0)return Int(1);
Int c=exp(b,n/2);
c=c*c;
if(n%2==1)return c*b;
return c;
}

int main()
{
int a,b;
cin>>a>>b;
Int A(a);
Int B(b);
A=exp(A,b);
B=exp(B,a);
// A.output();
// B.output();
if(A<B)
{
Int C=B-A;
cout<<"-"<<C.Int2Str()<<endl;
}
else
{
Int C=A-B;
// C.output();
// printf("%d\n",C.len);
cout<<C.Int2Str()<<endl;
}
return 0;
}