[training Day6] game [mathematics]

Ideas ：

We found the formula that can directly list the answers ：
$$ans=\sum _{i=1}^n\sum _{j=1}^m{H}_i\ast L_j\ast (m\ast (i-1)+j)$$
among H and L Represents the multiple of row and column expansion
Just row , delimit O（n+m） The formula that can be worked out is ok （ Row by yourself ）

$code$

#include<iostream>
#include<cstdio>

#define ll long long

using namespace std;

const ll MAXN = 1e6 + 10;
const ll mod = 1e9 + 7;

ll n, m, k;
ll l[MAXN], h[MAXN];

int main() {
    
	freopen("game.in", "r", stdin);
	freopen("game.out", "w", stdout);
	scanf("%lld%lld%lld", &n, &m, &k);
	for(ll i = 1; i <= m; i ++) l[i] = 1;
	for(ll i = 1; i <= n; i ++) h[i] = 1;
	for(ll i = 1; i <= k; i ++) {
    
		char ch; ll x, y;
		cin>>ch>>x>>y;
		if(ch == 'R') h[x] *= y, h[x] %= mod;
		else l[x] *= y, l[x] %= mod;
	}
	ll sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
	for(ll i = 1; i <= n; i ++) {
    
		sum2 = (sum2 + (i - 1) * m % mod * h[i] % mod) % mod;
		sum3 = (sum3 + h[i]) % mod;
	}
	for(ll i = 1; i <= m; i ++) {
    
		sum1 = (sum1 + l[i]) % mod;
		sum4 = (sum4 + i * l[i] % mod) % mod;
	}
	printf("%lld", (sum1 * sum2 % mod + sum3 * sum4 % mod) % mod);
}