loj2002(倍增)

题目链接

https://loj.ac/problem/2002

题解

感觉就是做 $n$ 次模卷积,由于模数比较小,所以直接倍增暴力就可以。。

对于要求至少一个质数先用上全部的减去用上非质数即可。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define mid (x+y>>1)
#define eps 1e-8
#define succ(x) (1<<(x))
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 100005
#define nm 20000005
using namespace std;
const double pi=acos(-1);
const ll inf=20170408;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}


int n,m,p,_m;
ll d[32][NM],ans[NM],_ans,tmp[NM];
bool v[nm];


int main(){
n=read();_m=m=read();p=read();
for(int i=2;i<=m;i++)if(!v[i])
for(int j=i<<1;j<=m;j+=i)v[j]++;
inc(i,0,p-1)d[0][i]=m/p;
m%=p;
inc(i,1,m)d[0][i]++;
m=p-1;
for(int i=1;succ(i)<=n;i++)
inc(j,0,m)inc(k,0,m)(d[i][(j+k)%p]+=d[i-1][j]*d[i-1][k])%=inf;
ans[0]=1;
for(int i=0;succ(i)<=n;i++)if(n>>i&1){
memcpy(tmp,ans,sizeof(ans));mem(ans);
inc(j,0,m)inc(k,0,m)(ans[(j+k)%p]+=tmp[j]*d[i][k])%=inf;
}
_ans=ans[0];
for(int i=2;i<=_m;i++)if(!v[i])d[0][i%p]--;
for(int i=1;succ(i)<=n;i++){
mem(d[i]);
inc(j,0,m)inc(k,0,m)(d[i][(j+k)%p]+=d[i-1][j]*d[i-1][k])%=inf;
}
mem(ans);ans[0]=1;
for(int i=0;succ(i)<=n;i++)if(n>>i&1){
memcpy(tmp,ans,sizeof(ans));mem(ans);
inc(j,0,m)inc(k,0,m)(ans[(j+k)%p]+=tmp[j]*d[i][k])%=inf;
}
_ans+=inf-ans[0];_ans%=inf;
printf("%lld\n",_ans);
return 0;
}