P5343 【XR-1】分块

 

 f[i ] += f[i-j]    j<=100  ( j belong to S

 

 

 构造的矩阵:

 

 

  这种求和的柿子可以化为矩阵乘法，做法是 矩阵对应元素设置0/1

#include <iostream>
#include <cstring>
#include <map>
using namespace std;
  const int N =1e6+10;
 #define int long long
  const int mod=1e9+7;
  map<int,bool> mp1,mp; 
 int n,m=10,f[N];
 
 // f[i]+=f[i-j] j->Set
 struct matrix {
    int a[105][105];
  };
 
  void init_(matrix &x){
     
    int i,j;
    for(i=1;i<=100;i++)
     for(j=1;j<=100;j++) {
         x.a[i][j]=0;
         if(i==j) x.a[i][j]=1;
     }
  }
  matrix mul(matrix &x,matrix &y){
    int i,j,k;
    matrix z;
     
    for(i=1;i<=100;i++)
     for(j=1;j<=100;j++){
        z.a[i][j]=0;
        for(k=1;k<=100;k++)
         z.a[i][j]+=x.a[i][k]*y.a[k][j], z.a[i][j]%=mod;
     }
     
    return z;
 }
   matrix ksm(matrix &x,int k){
    matrix tmp=x, ans;
    init_(ans);
     
    while(k){
        if(k&1) ans=mul(ans,tmp);
         
        tmp=mul(tmp,tmp);
        k/=2;
    }
    return ans;
 }
 signed main(){
 	cin>>n;
 	int i,j,t;
 	cin>>t;
 	while(t--){ int x; cin>>x; mp1[x]=1; }
 	cin>>t;
 	while(t--){ int x; cin>>x; if(mp1[x]) mp[x]=1; }
 	
 	matrix m0,m1;
 	memset(m0.a,0,sizeof m0);memset(m1.a,0,sizeof m1.a);
 	f[0]=1;
 	for(i=1;i<=100;i++)
	 	 for(j=1;j<=i;j++){
	 	 	if(mp[j]) f[i]+=f[i-j],f[i]%=mod;
	 	 }
 	if(n<100){
 		cout<<f[n]<<endl; return 0;
 	}
 	
 	
 	 for(j=1;j<=100;j++){
 	 	if(mp[j]==1) m0.a[1][j]=1;
 	 }
 	 for(i=2;i<=100;i++) m0.a[i][i-1]=1;
 	 for(i=1;i<=100;i++) m1.a[i][1]=f[100-i];
 	
    m0= ksm(m0,n-100+1);
 	m0= mul(m0,m1) ;
 	cout<<m0.a[1][1];
 }