[莫比乌斯反演 高斯消元 数学技巧] BZOJ 3601 一个人的数论

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;

const ll P=1e9+7;

inline ll Pow(ll a,ll b){
ll ret=1;
for (;b;b>>=1,a=a*a%P)
if (b&1)
ret=ret*a%P;
return ret;
}

inline ll Inv(ll a){
return Pow(a,P-2);
}

const int N=105;

int w,d;
ll ip[1005],ia[1005];
ll a

,A
;

int main(){
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
scanf("%d%d",&d,&w);
for (int i=1;i<=w;i++) scanf("%lld%lld",ip+i,ia+i);
ll isum=0;
for (int i=1;i<=d+2;i++){
ll tem=1;
for (int j=1;j<=d+2;j++) a[i][j]=tem,tem=tem*i%P;
a[i][d+3]=a[i-1][d+3]+Pow(i,d);
}
int m=d+2;
for (int i=1;i<=m;i++){
int k=i;
while (!a[k][i]) k++;
for (int j=1;j<=m+1;j++) swap(a[i][j],a[k][j]);
ll inv=Inv(a[i][i]);
for (int j=1;j<=m;j++){
if (j==i) continue;
ll t=a[j][i]*inv%P;
for (int k=1;k<=m+1;k++)
(a[j][k]+=P-a[i][k]*t%P)%=P;
}
}
for (int i=0;i<=d+1;i++)
A[i]=a[i+1][m+1]*Inv(a[i+1][i+1])%P;
ll Ans=0;
for (int i=0;i<=d+1;i++){
ll hi=1;
for (int j=1;j<=w;j++)
(hi*=(Pow(ip[j],ia[j]*i%(P-1))+P-Pow(ip[j],(d+(ia[j]-1)*i)%(P-1)))%P)%=P;
(Ans+=A[i]*hi%P)%=P;
}
printf("%lld\n",Ans);
return 0;
}