BZOJ4972 小Q的方格纸

　　每个格子记录其左下的45°直角梯形区域的和及左下矩形区域的和即可。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

#define ll long long

#define N 3010

#define M 3000010

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}

int gcd(int n,int m){return m==?n:gcd(m,n%m);}

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

int n,m,q;

unsigned int A,B,C,ans,p[M],a[N][N],b[N][N];

inline unsigned int rng61()

{

    A^=A<<;

    A^=A>>;

    A^=A<<;

    unsigned int t=A;

    A=B;

    B=C;

    C^=t^A;

    return C;

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj4972.in","r",stdin);

    freopen("bzoj4972.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    scanf("%d%d%d%u%u%u",&n,&m,&q,&A,&B,&C);

    for (int i=;i<=n;i++)

        for (int j=;j<=m;j++)

        a[i][j]=rng61();

    for (int i=n;i>=;i--)

        for (int j=;j<=m;j++)

        b[i][j]=a[i][j]+=a[i+][j];

    for (int i=n;i>=;i--)

        for (int j=;j<=m;j++)

        a[i][j]+=a[i+][j-],b[i][j]+=b[i][j-];

    p[]=;for (int i=;i<=q;i++) p[i]=p[i-]*;

    for (int i=;i<=q;i++)

    {

        int x=rng61()%n+,y=rng61()%m+,k=rng61()%min(x,y)+;

        unsigned int tot=a[x-k+][y]-a[x+][y-k]-b[x+][y]+b[x+][y-k];

        ans+=tot*p[q-i];

    }

    cout<<ans;

    return ;

}