【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=10;

int read()
{ 
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,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
");
        return re;
    }
    memset(re.data,0,sizeof(re.data));
    re.n = a.n; re.m = b.m;
    for(int i = 1; i <= a.n; i++)
    {
        for(int j = 1; j <= a.m; j++)
        {
            if(a.data[i][j] == 0) continue;
            for(int k = 1; 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
");
        return re;
    }
    re.n=re.m=a.n;
    memset(re.data,0,sizeof(re.data));
    for(int i=1;i<=re.n;i++) re.data[i][i]=1;
    while(b)
    {
        if(b&1) re=matrixMul(re,a); 
        a=matrixMul(a,a); 
        b>>=1;
    }
    return re;
}

void inputMat(int n,int m,matrix &a,ll *b)
{
    a.n = n; a.m = m;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            a.data[i][j] = *(b + (i - 1) * m + (j - 1));
}

void init(){
    ll pt[1][7] = {b,a,16,8,4,2,1};
    inputMat(1,7,ma,*pt);
    ll pt2[7][7] = {1,1,0,0,0,0,0,
                    2,0,0,0,0,0,0,
                    1,0,1,0,0,0,0,
                    4,0,4,1,0,0,0,
                    6,0,6,3,1,0,0,
                    4,0,4,3,2,1,0,
                    1,0,1,1,1,1,1,};
    inputMat(7,7,mb,*pt2);
}

int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%I64d%I64d%I64d",&n,&a,&b); 
        int i=3;
        a%=MOD; 
        b%=MOD;
        if(n == 1)
            printf("%I64d
",a);
        else if(n == 2)
            printf("%I64d
",b);
        else
        {
            
                init();
                ma=matrixMul(ma,matrixPow(mb,n-2));
        }
        printf("%I64d
",ma.data[1][1]);
    }
    return 0;
}