Problem:
Score : 100 points

Problem Statement

In K-city, there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?

Constraints

2 ≤ n, m ≤ 100

Input

Input is given from Standard Input in the following format:

n m

Output

Print the number of blocks in K-city.

Sample Input 1

3 4

Sample Output 1

6

There are six blocks, as shown below:

Sample Input 2

2 2

Sample Output 2

1

There are one block, as shown below:

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <tuple>
#include <cassert>
#include <functional>
#include <set>
#include <deque>
#include <list>
using namespace std;
typedef long long ll;
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
#define LNF __LONG_LONG_MAX__
const ll MOD = 1000000007;
const int MAX = 200000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// 法mでのaの逆元を計算
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
//最大公約数
long long gcd(long long x, long long y){
    if(y==0) return x;
    else return gcd(y,x%y);
}
//最小公倍数
long long lcm(long long x, long long y){
    return x / gcd(x, y) * y;
}
// nが大きく,kが10^7程度の時の二項係数
long long com[200005];
void combination(long long n,long long k){
    com[0] = 1;
    for(int i=1;i<=k;i++){
        com[i] = (com[i-1]*(n-i+1)%MOD)*modinv(i,MOD)%MOD;
    }
}

int main(){
    int n,m; cin >> n >> m;
    cout << (n-1)*(m-1) << endl;
    return 0;
}