AtCoder Beginner Contest 065（C

C - Reconciled?

设AA(x)为x的阶乘(对Mod求模)，n与m相差为2则输出0，相差为1则输出AA(n)*AA(n)*2%Mod,相同则输出AA(n)*AA(m)%Mod，这题好粗心，把ans变量写成int，改成ll就对了，可是比赛时硬是检查不出来5555555555555555

    #include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int Mod = 1e9+7;
    ll AA(ll x){
        ll ans = 1, i = 1;
        while(i <= x){
            ans = (ans*i)%Mod;
            i++;
        }
        return ans%Mod;
    }
    int main(){
        ll n,m;
        //cout << AA(100000) << endl;
        cin >> n >> m;
        if(abs(n-m) > 1){
            cout << 0 << endl;
            return 0;
        }
        if(abs(n-m) == 1){
            cout << (AA(n)*AA(m))%Mod << endl;
        }else {
            cout << ((AA(n)*2%Mod)*AA(m))%Mod << endl;
        }
        return 0;
    }