JZOJ6277矩阵游戏

Posted 2022-06-13 horizonwd

description

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analysis

• 设所有操作之后，\\(f[i]\\)表示\\(i\\)行乘的数，\\(g[j]\\)表示\\(j\\)列乘的数，那么

\\[Answer=\\sum^n_i=1\\sum^m_j=1[m*(i-1)+j]*f[i]*g[j]\\]

• 中括号里的就是该位置原来的数，很好理解，然后移项

\\[=\\sum_i=1^nf[i]\\sum_j=1^mg[j]*m*(i-1)+g[j]*j\\]

\\[=\\sum_i=1^n\\f[i]*(i-1)*m*\\sum_j=1^mg[j]+f[i]*\\sum_j=1^mg[j]*j\\\\]

• 然后就没了

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code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 1000005
#define mod 1000000007
#define mo mod
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[MAXN],g[MAXN];
ll n,m,k,sigma,sum,ans;
char s[5];

inline ll read()

    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch)if (ch=='-')f=-1;ch=getchar();
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;

int main()

    freopen("T1.in","r",stdin);
    //freopen("game.in","r",stdin);
    //freopen("game.out","w",stdout);
    n=read(),m=read(),k=read();
    fo(i,1,n)f[i]=1;fo(i,1,m)g[i]=1;
    while (k--)
    
        scanf("%s",&s);
        ll x=read(),y=read();scanf("\\n");
        if (s[0]=='R')f[x]=(f[x]*y)%mod;
        else g[x]=(g[x]*y)%mod;
    
    fo(i,1,m)sigma=(sigma+g[i])%mod,sum=(sum+g[i]*i)%mod;
    fo(i,1,n)ans=(ans+(f[i]%mod*(i-1)%mod*m%mod*sigma%mod+f[i]%mod*sum%mod)%mod)%mod;
    printf("%lld\\n",ans);
    return 0;