【五校联考2day2】WYF的盒子

Description

求

∑i=mnikmodp

90% n,m,p<=10^12,k<=2000
另外10% n-m<=5000,n,m,p,k<=10^12

Solution

采用自然数幂和
注意，这种东西的模数比较大，两个乘起来会爆longlong。
于是可以用乘法取模黑科技
（懒癌发作ing~）

Code

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define fo(i,a,b) for(ll i=a;i<=b;i++)#define N 2005#define maxn 10000000#define ll long longusing namespace std;ll s[N][N],f[N],n,m,p,k;ll sum(ll x,ll y){    ll a1=x/maxn,a2=x%maxn,b1=y / maxn,b2=y%maxn;    ll t=a1*b1%p*maxn%p*maxn%p;    t=(t+a2*b2%p);    t=(t+a1*b2%p*maxn%p);    t=(t+a2*b1%p*maxn%p);    return t%p;}ll mi(ll x,ll y) {    if (y==1) return x;    ll z=mi(x,y / 2);    z=sum(z,z);    if (y%2) z=sum(z,x);    return z;}ll calc(ll n){    f[0]=n%p;    if (n%2) f[1]=sum(n,(n+1)/2);    else f[1]=sum(n/2,n+1);    fo(i,2,k){        f[i]=1;        fo(j,n+1-i,n+1)            if (j%(i+1)==0) f[i]=sum(f[i],j/(i+1));else f[i]=sum(f[i],j);        fo(j,0,i-1) f[i]=(f[i]-sum(s[i][j],f[j])+p)%p;    }    return f[k];}void per() {    fo(i,0,k) s[i][0]=0,s[i][i]=1;    fo(i,1,k)        fo(j,1,i-1)            s[i][j]=(s[i-1][j-1]-(i-1)*s[i-1][j]%p+p)%p;}int main() {    scanf("%lld%lld%lld%lld",&k,&n,&m,&p);    if (k>2000) {        ll ans=0;fo(i,m,n) ans=(ans+mi(i,k)%p)%p;        printf("%lld",ans);return 0;        }    per();    printf("%lld",(calc(n)-calc(m-1)+p)%p);}