sgu112

maksyuki 发表于 oj 分类，标签: 基础算法-高精度

27 8月 2015

a^b-b^a

You are given natural numbers a and b. Find a^b-b^a.

Input

Input contains numbers a and b (1≤a,b≤100).

Output

Write answer to output.

Sample Input

2 3

Sample Output

-1

题目类型：高精度运算

算法分析：由于指数比较小，所以可以直接使用高精度乘法来模拟

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 100000 + 66;

string Mul (string a, string b)//高精度乘法a,b,均为非负整数
{
    string s;// maxn表示相乘的数的最大可能位数
    int na[maxn], nb[maxn], nc[maxn], La = a.size (), Lb = b.size ();//na存储被乘数，nb存储乘数，nc存储积

    fill (na, na + maxn, 0); fill (nb, nb + maxn, 0); fill (nc, nc + maxn, 0);//将na,nb,nc都置为0

    for(int i = La - 1; i >= 0; i--)
        na[La-i] = a[i] - '0';//将字符串表示的大整形数转成i整形数组表示的大整形数

    for (int i = Lb - 1; i >= 0; i--)
        nb[Lb-i] = b[i] - '0';

    for (int i = 1; i <= La; i++)
        for (int j = 1; j <= Lb; j++)
            nc[i+j-1] += na[i] * nb[j];//a的第i位乘以b的第j位为积的第i+j-1位（先不考虑进位）

    for (int i = 1; i <= La + Lb; i++)
    {
        nc[i+1] += nc[i] / 10;
        nc[i] %= 10;//统一处理进位
    }

    if (nc[La+Lb])
        s += nc[La+Lb] + '0';//判断第i+j位上的数字是不是0

    for (int i = La + Lb - 1; i >= 1; i--)
        s += nc[i] + '0';//将整形数组转成字符串

return s;
}

string Sub (string a, string b)//适用于两个非负整数的相减
{
    bool is_neg = false;//表示结果的正负
    if ((a.size () < b.size ()) || (a.size () == b.size () && a < b))
    {
        is_neg = true;
        swap (a, b);
    }

    string ans;
    int na[maxn] = {0}, nb[maxn] = {0};// maxn表示两个相减的数的位数

    int la = a.size (), lb = b.size ();

    for (int i = 0; i < la; i++)
        na[la-1-i] = a[i] - '0';

    for (int i = 0; i < lb; i++)
        nb[lb-1-i] = b[i] - '0';

    int lmax = la > lb ? la : lb;

    for (int i = 0; i < lmax; i++)
    {
        na[i] -= nb[i];
        if (na[i] < 0)
        {
            na[i] += 10;
            na[i+1]--;
        }
    }

    while (!na[--lmax] && lmax > 0) ;

    lmax++;

    if (is_neg)
        ans += "-";

    for (int i = lmax - 1; i >= 0; i--)
        ans += na[i] + '0';

return ans;
}

int main()
{
//	ifstream cin ("aaa.txt");
//	freopen ("aaa.txt", "r", stdin);

    string a, b;
	cin >> a >> b;

	int va = 0, vb = 0;
    for (int i = 0; a[i]; i++)
		va = 10 * va + (a[i] - '0');

	for (int i = 0; b[i]; i++)
		vb = vb * 10 + (b[i] - '0');


	string sa = a, sb = b;

    for (int i = 1; i <= vb - 1; i++)
        sa = Mul(sa, a);

    for (int i = 1; i <= va - 1; i++)
		sb = Mul (sb, b);

   cout << Sub (sa, sb) << endl;

return 0;
}

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#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 100000 + 66;

string Mul (string a, string b)//高精度乘法a,b,均为非负整数

{

string s;// maxn表示相乘的数的最大可能位数

int na[maxn], nb[maxn], nc[maxn], La = a.size (), Lb = b.size ();//na存储被乘数，nb存储乘数，nc存储积

fill (na, na + maxn, 0); fill (nb, nb + maxn, 0); fill (nc, nc + maxn, 0);//将na,nb,nc都置为0

for(int i = La - 1; i >= 0; i--)

na[La-i] = a[i] - '0';//将字符串表示的大整形数转成i整形数组表示的大整形数

for (int i = Lb - 1; i >= 0; i--)

nb[Lb-i] = b[i] - '0';

for (int i = 1; i <= La; i++)

for (int j = 1; j <= Lb; j++)

nc[i+j-1] += na[i] * nb[j];//a的第i位乘以b的第j位为积的第i+j-1位（先不考虑进位）

for (int i = 1; i <= La + Lb; i++)

{

nc[i+1] += nc[i] / 10;

nc[i] %= 10;//统一处理进位

}

if (nc[La+Lb])

s += nc[La+Lb] + '0';//判断第i+j位上的数字是不是0

for (int i = La + Lb - 1; i >= 1; i--)

s += nc[i] + '0';//将整形数组转成字符串

return s;

}

string Sub (string a, string b)//适用于两个非负整数的相减

{

bool is_neg = false;//表示结果的正负

if ((a.size () < b.size ()) || (a.size () == b.size () && a < b))

{

is_neg = true;

swap (a, b);

}

string ans;

int na[maxn] = {0}, nb[maxn] = {0};// maxn表示两个相减的数的位数

int la = a.size (), lb = b.size ();

for (int i = 0; i < la; i++)

na[la-1-i] = a[i] - '0';

for (int i = 0; i < lb; i++)

nb[lb-1-i] = b[i] - '0';

int lmax = la > lb ? la : lb;

for (int i = 0; i < lmax; i++)

{

na[i] -= nb[i];

if (na[i] < 0)

{

na[i] += 10;

na[i+1]--;

}

while (!na[--lmax] && lmax > 0) ;

lmax++;

if (is_neg)

ans += "-";

for (int i = lmax - 1; i >= 0; i--)

ans += na[i] + '0';

return ans;

}

int main()

{

// ifstream cin ("aaa.txt");

// freopen ("aaa.txt", "r", stdin);

string a, b;

cin >> a >> b;

int va = 0, vb = 0;

for (int i = 0; a[i]; i++)

va = 10 * va + (a[i] - '0');

for (int i = 0; b[i]; i++)

vb = vb * 10 + (b[i] - '0');

string sa = a, sb = b;

for (int i = 1; i <= vb - 1; i++)

sa = Mul(sa, a);

for (int i = 1; i <= va - 1; i++)

sb = Mul (sb, b);

cout << Sub (sa, sb) << endl;

return 0;

}

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