There is an infinite tree with roots. Each node has N children. The distance between each child and the parent node is di. Find the number of nodes whose distance from the root node is less than or equal to X

Idea: pay attention to the observation data range, each d[i] is less than or equal to 100, so we can set dp[i] to represent the number of points from the origin I, sum[i] to represent the total, and finally sum[x] to calculate.

State transfer equation

1 for(int i=1;i<=100;i++)
2     {
3         for(int j=1;j<=i;j++)
4         {
5             dp[i]=(dp[i]+dp[i-j]*cnt[j])%MOD;
6         }
7         sum[i]=(sum[i-1]+dp[i])%MOD;
8     }

But because x is too large, and dp[n+1] is every dp[i]*cnt[n+1-i], let N=100, save the first 100 numbers of n+1, update sum with the 101st column, and solve it with the fast power of matrix. (C is the unit matrix!!!)

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 const ll MOD=(ll)1e9+7;
  5 
  6 struct matrix
  7 {
  8     ll x,y;
  9     ll a[112][112];
 10     matrix(){
 11     };
 12     matrix(ll xx,ll yy):x(xx),y(yy)
 13     {
 14         memset(a,0,sizeof(a));
 15     }
 16 }Base;
 17 
 18 matrix mul (matrix A,matrix B)
 19 {
 20     matrix C(A.x,B.y);
 21     for(int i=1;i<=A.x;i++)
 22     {
 23         for(int j=1;j<=B.y;j++)
 24         {
 25             for(int k=1;k<=B.x;k++)
 26             {
 27                 C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j])%MOD;
 28             }
 29         }
 30     }
 31     return C;
 32 }
 33 
 34 ll n,x,cnt[120],sum[120],dp[120];
 35 void init()
 36 {
 37     memset(cnt,0,sizeof(cnt));
 38     scanf("%lld%lld",&n,&x);
 39     for(int i=1;i<=n;i++)
 40     {
 41         int p;
 42         scanf("%d",&p);
 43         cnt[p]++;
 44     }
 45     memset(sum,0,sizeof(sum));
 46     dp[0]=1; sum[0]=1;
 47     for(int i=1;i<=100;i++)
 48     {
 49         for(int j=1;j<=i;j++)
 50         {
 51             dp[i]=(dp[i]+dp[i-j]*cnt[j])%MOD;
 52         }
 53         sum[i]=(sum[i-1]+dp[i])%MOD;
 54     }
 55     
 56     Base.x=101; Base.y=101;
 57     //123...97 98 99  100 101(ans)
 58     
 59     //000...000 len[100] len[100]
 60     //100...000 len[99] len[99]
 61     //010...000 len[98] len[98]
 62     //.........................
 63     //000...001 len[1] len[1]
 64     //000...000 0        1
 65     for(int i=1;i<=99;i++)
 66     {
 67         Base.a[i+1][i]=1;
 68         Base.a[i][100]=cnt[101-i];
 69         Base.a[i][101]=cnt[101-i];
 70     }
 71     Base.a[100][100]=Base.a[100][101]=cnt[1];
 72     Base.a[101][101]=1;
 73 }
 74 
 75 matrix qpow (matrix Base,ll b)
 76 {
 77     matrix C(Base.x,Base.y);
 78     for(int i=1;i<=101;i++) C.a[i][i]=1;//danweijuzhen
 79     while(b)
 80     {
 81         if(b%2==1) C=mul(C,Base);
 82         b/=2;
 83         Base=mul(Base,Base);
 84     }
 85     return C;
 86 }
 87 
 88 int main()
 89 {
 90     init();
 91     matrix A(1,101);
 92     for(int i=1;i<=100;i++) A.a[1][i]=dp[i];
 93     A.a[1][101]=sum[100];
 94     if(x<=100)
 95     {
 96         printf("%lld\n",sum[x]);
 97         return 0;
 98     }
 99     A=mul(A,qpow(Base,x-100));
100     printf("%lld\n",A.a[1][101]%MOD);
101     return 0;
102 }