答案如下，如果不会的可以看看哦

#include <stdio.h>
#define inf 1999999999
#define md 1000000007
#define min(a,b) a<b?a:b
#define max(a,b) a>b?a:b
int az[12],al[12],ar[12],w[12],aa[500010][12];
int B[12],C[500010],D[500010],la[500010][12],ra[500010][12];
int ksm(int a,int b) {
int jg=1;
while(b>0) {
if(b&1)
jg=1lljga%md;
a=1llaa%md;
b=(b>>1);
}
return jg;
}
int Cc[12][12],xs[12][12];
void GetB(int k)//伯努利数
{
B[0]=1;
for (int i=1;i<=k+1;i++) {
for (int j=0;j<=i;j++) {
if(j0||ji)Cc[i][j]=1;
else Cc[i][j]=(Cc[i-1][j]+Cc[i-1][j-1])%md;
}
}
for (int i=1;i<=k;i++) {
int h=0;
for (int j=0;j<i;j++)
h=(h+1llCc[i+1][j]B[j])%md;
B[i]=1ll(md-h)ksm(i+1,md-2)%md;
}
for (int i=1;i<=k;i++) {
int ny=ksm(i+1,md-2);
for (int j=0;j<=i;j++)
xs[i][i+1-j]=1llCc[i+1][j]B[j]%mdny%md;
}
}
struct SLi//一次函数
{
int k,b;
SLi() {}
SLi(int K,int B) {
k=K;b=B;
}
SLi(int Z) {
k=0;b=Z;
}
};
SLi operator-(const SLi &x,const SLi &y) {
return SLi(x.k-y.k,x.b-y.b);
}
int Sum(int k,int n) {
if(k==0)
return n+1;
int jg=0;
for (int i=0,j=1;i<=k+1;i++) {
jg=(jg+1lljxs[k][i])%md;
j=1llj*(n+1)%md;
}
return jg;
}
struct DXS//多项式
{
int sz[12],n;
void operator=(const DXS &a) {
n=a.n;
for (int i=0;i<=n;i++)
sz[i]=a.sz[i];
}
void clear() {
for (int i=1;i<=10;i++)
sz[i]=0;
sz[0]=1;n=0;
}
int sum(int l,int r) {
int ans=0;
for (int i=0;i<=n;i++) {
int t=(Sum(i,r)-Sum(i,l-1)+md)%md;
ans=(ans+1llsz[i]t)%md;
}
return ans;
}
};
DXS operator(const DXS&x,const SLi&y) {
DXS rt;
rt.n=x.n+1;
rt.sz[rt.n]=0;
for (int i=0;i<=x.n;i++)
rt.sz[i]=1lly.bx.sz[i]%md;
for (int i=0;i<=x.n;i++)
rt.sz[i+1]=(rt.sz[i+1]+1lly.kx.sz[i])%md;
return rt;
}
struct SQj//维护区间
{
int l,r;
SQj() {}
SQj(int L,int R) {
l=L;r=R;
}
};
SQj jiao(const SQj&a,const SQj&b) {
return SQj(max(a.l,b.l),min(a.r,b.r));
}
int floor(int,int);
int ceil(int x,int y) {
if(y<0)x=-x,y=-y;
if(x>=0)return (x+y-1)/y;
return -floor(-x,y);
}
int floor(int x,int y) {
if(y<0)x=-x,y=-y;
if(x>=0)return x/y;
return -ceil(-x,y);
}
SQj Less(SLi a,SLi b) {
int x=a.k-b.k,y=b.b-a.b;
if(x0)return y>=0?SQj(-inf,inf):SQj(inf,-inf);
if(x>0)return SQj(-inf,floor(y,x));
return SQj(ceil(y,x),inf);
}
SQj More(SLi a,SLi b) {
int x=a.k-b.k,y=b.b-a.b;
if(x0)return y<=0?SQj(-inf,inf):SQj(inf,-inf);
if(x>0)return SQj(ceil(y,x),inf);
return SQj(-inf,floor(y,x));
}
int cal0(int n,int i,int k) {
int l[12],r[12],jg=1;
for (int c=1;c<=k;c++)
l[c]=la[i][c],r[c]=ra[i][c];
for (int c=1;c<=k;c++) {
int zl=1-l[c],zr=w[c]-r[c];
if(c!=C[(i+1)%n]) {
int s=zr-zl+1;
if(s<0)s=0;
jg=1lljgs%md;
}
else {
int t=aa[i][c],s=0;
if(D[(i+1)%n]==1) {
int o=w[c]-t;
if(o>=zl&&o<=zr)s=1;
}
else {
int o=1-t;
if(o>=zl&&o<=zr)s=1;
}
jg=1lljgs%md;
}
}
return 1ll(i+2)jg%md;
}
int main() {
int n,k;
scanf("%d%d",&n,&k);GetB(k);
for (int i=1;i<=k;i++)
scanf("%d",&w[i]);
for (int i=0;i<n;i++) {
scanf("%d%d",&C[i],&D[i]);
if(i>0) {
for (int j=1;j<=k;j++)
aa[i][j]=aa[i-1][j];
}
aa[i][C[i]]+=D[i];//循环节中某一前缀的偏移量
az[C[i]]+=D[i];
if(az[C[i]]<al[C[i]])//最左移位
al[C[i]]=az[C[i]];
if(az[C[i]]>ar[C[i]])//最右移位
ar[C[i]]=az[C[i]];
for (int j=1;j<=k;j++) {
la[i][j]=al[j];ra[i][j]=ar[j];
}
}
bool zd=false;
for (int i=1;i<=k;i++) {
if(az[i]!=0||ar[i]-al[i]>=w[i]) {
zd=true;break;
}
}
if(!zd)//走不出去
{
printf("-1");
return 0;
}
int ans=1;
for (int i=1;i<=k;i++) {
if(i!=C[0])
ans=1llansw[i]%md;
}
for (int i=0;i<n;i++) {
ans=(ans+cal0(n,i,k))%md;//特殊处理x=0的情况
SLi l[12],r[12],d[12];
for (int c=1;c<=k;c++)//算出对应维度的一次函数
{
if(az[c]>=0) {
l[c]=al[c];
int t=az[c]+ra[i][c];
if(ar[c]>t)t=ar[c];
r[c]=SLi(az[c],t-az[c]);
}
else {
r[c]=ar[c];
int t=az[c]+la[i][c];
if(al[c]<t)t=al[c];
l[c]=SLi(az[c],t-az[c]);
}
d[c]=r[c]-l[c];
}
int tc=C[(i+1)%n];
SLi o;SLi tz=SLi(az[tc],aa[i][tc]);
if(D[(i+1)%n]==1)o=w[tc]-tz;
else o=1-tz;
SLi zl=1-l[tc],zr=w[tc]-r[tc];//tc这一维度起点的范围
SQj qj=jiao(More(o,zl),Less(o,zr));//tc这一维度起点是确定的，需要满足条件
for (int i=1;i<=k;i++) {
d[i]=w[i]-d[i];
qj=jiao(qj,More(d[i],1));//方案数>0
}
qj=jiao(qj,SQj(1,inf));
if(qj.l>qj.r)continue;
DXS ji;ji.clear();ji=jiSLi(n,i+2);
for (int c=1;c<=k;c++)
{
if(c!=tc)
ji=ji*d[c];
}
ans=(ans+ji.sum(qj.l,qj.r))%md;
}
printf("%d",(ans%md+md)%md);
return 0;
}