题解2

#include<bits/stdc++.h>
#define ri register int
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int N=2e6+3;
ll ans=0;
ll jie[N],ni[N];
struct node
{
int x,y,fl;
void init(int a,int b,int c)
{
x=a,y=b,fl=c;
}
}p[20];
int tot;
int x3,y3,x4,y4;
int xx[10],yy[10];
ll qm(ll x,ll y){
ll ret=1;
while(y){
if(y&1) ret=retx%mod;
x=xx%mod;
y>>=1;
}
return ret;
}
ll C(int n,int m)
{
return jie[n]ni[m]%modni[n-m]%mod;
}
ll G(int x1,int y1,int x2,int y2)
{
ll A=abs(x2-x1),B=abs(y2-y1);
return C(A+B,B);
}
ll sol(int x1,int y1,int x2,int y2,int f1,int f2)
{
ll ret=0;
for(ri x=x3;x<=x4;++x){
ret=(ret+G(x1,y1,x,y4)G(x,y4+1,x2,y2)%mod(x+y4+1))%mod;//注意这里的x+y4+1的+1
ret=(ret-G(x1,y1,x,y3-1)G(x,y3,x2,y2)%mod(x+y3)%mod+mod)%mod;
}
for(ri y=y3;y<=y4;++y)
{
ret=(ret+G(x1,y1,x4,y)G(x4+1,y,x2,y2)%mod(x4+y+1))%mod;
ret=(ret-G(x1,y1,x3-1,y)G(x3,y,x2,y2)%mod(x3+y)%mod+mod)%mod;
}
ret=ret*(f1*f2);

return ret;

}
int main()
{
int x,y;
for(ri i=1;i<=6;++i)scanf("%d",&xx[i]);
for(ri i=1;i<=6;++i)scanf("%d",&yy[i]);
jie[0]=1;
for(ri i=1;i<=N-2;++i) jie[i]=jie[i-1]i%mod;
ni[N-2]=qm(jie[N-2],mod-2);
for(ri i=N-3;i>=0;--i) ni[i]=ni[i+1](i+1)%mod;

x3=xx[3],y3=yy[3],x4=xx[4],y4=yy[4];
p[1].init(xx[1]-1,yy[1]-1,1);p[2].init(xx[2],yy[1]-1,-1);//注意这里是这样的 
p[3].init(xx[1]-1,yy[2],-1);p[4].init(xx[2],yy[2],1);

p[5].init(xx[5],yy[5],1);p[6].init(xx[6]+1,yy[5],-1);
p[7].init(xx[5],yy[6]+1,-1);p[8].init(xx[6]+1,yy[6]+1,1);

for(int i=1;i<=4;++i)

{
for(int j=5;j<=8;++j)
{
ans=(ans+sol(p[i].x,p[i].y,p[j].x,p[j].y,p[i].fl,p[j].fl)+mod)%mod;
}
}
printf("%lld",ans);
return 0;
}