T1 81101noip simulation competition T1

Tag: getchar score + = STD-optimized RET fine space GETC

 

Ideas:

We can see this question, and we can think of one-dimensional difference at a glance.

However, this complexity is O (NQ), and obviously t

So how to optimize it?

We will find that in the difference ~ Within the L-1 range

The values added by the difference are the same as the X-axis, and the y-axis increases progressively.

The abscissa and ordinate values of the reduced values increase with 1 as the tolerances.

Then, we can divide the difference Array

Mark (R, c) (R, C + 1), (R + L, c) (R + L, C + l) each time.

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define rii register int i#define rij register int j#define int long long using namespace std;int cf1[2005][2005],cf2[2005][2005],x[2005][2005];int n,q,r,c,l,s;inline int rd(){    int y=0,f=1;char ch=getchar();    while(!isdigit(ch)) {f=ch==‘-‘?0:1;ch=getchar();}    while(isdigit(ch))  {y=(y<<1)+(y<<3)+ch-‘0‘;ch=getchar();}    return f?y:-y;}signed main(){    freopen("u.in","r",stdin);    freopen("u.out","w",stdout);    n=rd(),q=rd();    for(rii=1;i<=q;i++)    {        r=rd(),c=rd(),l=rd(),s=rd();        cf1[r][c]+=s;        cf1[r+l][c]-=s;        cf2[r][c+1]+=s;        cf2[r+l][c+l+1]-=s;    }    for(rij=1;j<=n;j++)    {        for(rii=1;i<=n;i++)        {            cf1[i][j]+=cf1[i-1][j];        }    }    for(rii=1;i<=n;i++)    {        for(rij=1;j<=n;j++)        {            cf2[i][j]+=cf2[i-1][j-1];        }    }    for(rii=1;i<=n;i++)    {        for(rij=1;j<=n;j++)        {            x[i][j]+=x[i][j-1]+cf1[i][j]-cf2[i][j];        }    }    int ans=0;    for(rii=1;i<=n;i++)    {        for(rij=1;j<=n;j++)        {            ans^=x[i][j];        }    }    printf("%lld",ans);    return 0;}

 

T1 81101noip simulation competition T1