112. \(a^b-b^a\)

time limit per test: 0.25 sec / memory limit per test: 4096 KB

You are given natural numbers a and b. Find ab-ba.

Input

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

Output

Write answer to output.

Example(s)

Sample Input	Sample Output
2 3	-1

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <ctime>
#include <map>

#define inf 0x3f3f3f3f
#define Inf 0x3FFFFFFFFFFFFFFFLL
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define Rep(i, n) for (int i = 1; i <= (n); ++i)
#define clr(x, a) memset(x, (a), sizeof x)
using namespace std;
typedef long long ll;

struct BN {
#define BN_L 9
#define BN_N 1100
#define BN_M 1000000000
  ll c[BN_N];
  int n;

  BN() {
    clr(c, 0);
    n = 0;
    c[n++] = 0;
  }

  BN(ll num) {
    clr(c, 0);
    n = num > 0 ? 0 : 1;
    while (num > 0) {
      c[n++] = num % BN_M;
      num /= BN_M;
    }
  }

  friend BN operator+(BN a, BN b) {
    BN t = BN();
    int l = max(a.n, b.n);
    rep(i, l) {
      t.c[i] += (a.c[i] + b.c[i]);
      if (t.c[i] >= BN_M) {
        t.c[i] -= BN_M;
        t.c[i + 1] += 1;
      }
    }
    if (t.c[l] > 0) {
      t.n = l + 1;
    } else {
      t.n = l;
    }
    return t;
  }

  // assume a >= b
  friend BN operator-(BN a, BN b) {
    BN t = BN();
    int l = max(a.n, b.n);
    rep(i, l) {
      t.c[i] += (a.c[i] - b.c[i]);
      if (t.c[i] < 0) {
        t.c[i] += BN_M;
        t.c[i + 1] -= 1;
      }
    }
    while (t.c[l] == 0) --l;
    t.n = max(l + 1, 1);
    return t;
  }

  friend BN operator*(BN a, BN b) {
    BN t = BN();
    int l = a.n + b.n;
    rep(i, l) {
      rep(j, i + 1) {
        t.c[i] += a.c[j] * b.c[i - j];
        t.c[i + 1] += t.c[i] / BN_M;
        t.c[i] %= BN_M;
      }
    }
    while (t.c[l] == 0) --l;
    t.n = max(l + 1, 1);
    return t;
  }

  friend BN operator^(BN a, ll m) {
    BN t = BN(1);
    for (; m > 0; m >>= 1) {
      if (m & 1) t = t * a;
      a = a * a;
    }
    return t;
  }

  friend bool operator==(BN a, BN b) {
    int l = max(a.n, b.n);
    rep(i, l) {
      if (a.c[i] != b.c[i]) {
        return 0;
      }
    }
    return 1;
  }

  friend bool operator<(BN a, BN b) {
    int l = max(a.n, b.n);
    for (int i = l - 1; i >= 0; --i) {
      if (a.c[i] < b.c[i]) {
        return 1;
      } else if (a.c[i] > b.c[i]) {
        return 0;
      }
    }
    return 0;
  }

  void pr() {
    for (int i = n - 1; i >= 0; --i) {
      printf(i == n - 1 ? "%I64d" : "%09I64d", c[i]);
    }
    putchar('\n');
  }
};

BN a, b, c;

int main() {
  int x, y; scanf("%d%d", &x, &y);
  a = BN(x) ^ y;
  b = BN(y) ^ x;
  if (a < b) {
    putchar('-');
    (b - a).pr();
  } else {
    (a - b).pr();
  }
  return 0;
}