模拟13 题解（waiting）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #define R register
 6 #define int long long
 7 using namespace std;
 8 int read()
 9 {
10     int f=1,x=0;char ch=getchar();
11     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
12     while(ch<=‘9‘&&ch>=‘0‘){x=(x<<3)+(x<<1)+ch-‘0‘;ch=getchar();}
13     return f*x;
14 }
15 const int maxn=1000005;
16 const int mod=1e9+7;
17 int n,m,k;
18 int a[maxn],b[maxn];
19 int fa[maxn],fb[maxn];
20 char opt[3];
21 int cal(int x,int y)
22 {
23     return ((x-1)*m%mod+y)%mod;
24 }
25 signed  main()
26 {
27 //    freopen("data","r",stdin);
28     n=read(),m=read(),k=read();
29     for(int i=1;i<=n;i++)
30         fa[i]=1;
31     for(int i=1;i<=m;i++)
32         fb[i]=1;
33     while(k--)
34     {
35         scanf("%s",opt);
36         R int x=read(),y=read();
37         if(opt[0]==‘R‘)
38         {
39             fa[x]=fa[x]*y%mod;
40         }
41         else 
42         {
43             fb[x]=fb[x]*y%mod;
44         }
45     }
46     int sum=0,dtb=0,ta=0;
47     for(int i=1;i<=n;i++){
48         ta=(ta+(i-1)*m%mod*fa[i]%mod)%mod;
49         dtb=(dtb+fa[i])%mod;
50     }
51     for(int j=1;j<=m;j++)
52         (sum+=fb[j]*ta%mod+j*fb[j]%mod*dtb%mod)%=mod;;
53     printf("%lld\n",sum%mod);
54 }