【HDOJ5950】Recursive sequence（矩阵乘法，快速幂）

题意：f[1]=a,f[2]=b,f[i]=2f[i-2]+f[i-1]+i^4（i>=3），多组询问求f[n]对2147493647取模

N,a,b < 2^31

思路：重点在于i^4的处理

对于i转移矩阵中可以记录下它的0,1,2,3,4次项

i的幂又可以由i-1的幂运算得出，最后推出的系数是二项式展开的系数

试试新的矩乘模板

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   2100000
 #define MOD 2147493647
 #define eps 1e-8
 #define pi acos(-1)
 const int MAXN=;
 
 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
 
 struct matrix       //矩阵类
 {
     int n,m;
     ll data[MAXN][MAXN];
 };
 
 matrix ma,mb;
 ll a,b,c,d,p,n;
 
 matrix matrixMul(matrix a, matrix b)
 {
     matrix re;
     if(a.m!=b.n)
     {
         printf("error\n");
         return re;
     }
     memset(re.data,,sizeof(re.data));
     re.n = a.n; re.m = b.m;
     for(int i = ; i <= a.n; i++)
     {
         for(int j = ; j <= a.m; j++)
         {
             if(a.data[i][j] == ) continue;
             for(int k = ; k <= b.m; k++)
             {
                 re.data[i][k] += (a.data[i][j] % MOD * b.data[j][k] % MOD) % MOD;
                 re.data[i][k] %= MOD;
             }
         }
     }
     return re;
 }
 
 matrix matrixPow(matrix a,int b)
 {
     matrix re;
     if(a.n!=a.m)
     {
         printf("error2\n");
         return re;
     }
     re.n=re.m=a.n;
     memset(re.data,,sizeof(re.data));
     for(int i=;i<=re.n;i++) re.data[i][i]=;
     while(b)
     {
         if(b&) re=matrixMul(re,a);
         a=matrixMul(a,a);
         b>>=;
     }
     return re;
 }
 
 void inputMat(int n,int m,matrix &a,ll *b)
 {
     a.n = n; a.m = m;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= m; j++)
             a.data[i][j] = *(b + (i - ) * m + (j - ));
 }
 
 void init(){
     ll pt[][] = {b,a,,,,,};
     inputMat(,,ma,*pt);
     ll pt2[][] = {,,,,,,,
                     ,,,,,,,
                     ,,,,,,,
                     ,,,,,,,
                     ,,,,,,,
                     ,,,,,,,
                     ,,,,,,,};
     inputMat(,,mb,*pt2);
 }
 
 int main()
 {
     int cas;
     scanf("%d",&cas);
     while(cas--)
     {
         scanf("%I64d%I64d%I64d",&n,&a,&b);
         int i=;
         a%=MOD;
         b%=MOD;
         if(n == )
             printf("%I64d\n",a);
         else if(n == )
             printf("%I64d\n",b);
         else
         {
 
                 init();
                 ma=matrixMul(ma,matrixPow(mb,n-));
         }
         printf("%I64d\n",ma.data[][]);
     }
     return ;
 }