hdu5950

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and
i
4
. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

题意：

：F(n) = 2*F(n-2) + F(n-1) + n4 和F(1) = a，F(2) = b；给定一个N，求F(N)
再来一道矩阵快速幂吧~~出现n4考虑（（n-1）+1）4。。。。二项式定理。。。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const long long MOD = 2147493647;struct Matrix{    long long m[7][7];};long long aa,bb,n;Matrix Mul(Matrix a ,Matrix b){    Matrix tmp;    memset(tmp.m,0,sizeof(tmp.m));    for(int i =  0 ; i < 7 ; i++)        for(int j = 0 ; j < 7 ;j++)            for(int k = 0 ; k < 7 ; k++)                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j])%MOD;    return tmp;}Matrix qpow(Matrix a,long long n){    Matrix tmp;    memset(tmp.m,0,sizeof(tmp.m));    for(long long i = 0 ; i < 7 ; i++)  tmp.m[i][i] = 1;        while(n){            if(n & 1)   tmp = Mul(tmp,a);            n >>= 1;            a = Mul(a,a);        }    return tmp;}void sov(){    Matrix tmp;    memset(tmp.m,0,sizeof(tmp.m));    tmp.m[0][0] = tmp.m[0][2] = tmp.m[0][6] = tmp.m[1][0] = 1;    for(int i = 2 ; i <= 6 ; i++)   tmp.m[i][6] = tmp.m[i][i] = 1;    tmp.m[0][3] = tmp.m[0][5] = tmp.m[2][3] = tmp.m[2][5] = 4;    tmp.m[0][4] = tmp.m[2][4] = 6;    tmp.m[3][5] = tmp.m[3][4] = 3;    tmp.m[4][5] = tmp.m[0][1] = 2;    tmp = qpow(tmp,n-2);    long long ans;    ans = (tmp.m[0][0] * bb)%MOD;    ans = (ans+tmp.m[0][1] *aa)%MOD;    ans = (ans+tmp.m[0][2] *16)%MOD;    ans = (ans+tmp.m[0][3] * 8)%MOD;    ans = (ans+tmp.m[0][4] * 4)%MOD;    ans = (ans+tmp.m[0][5] * 2)%MOD;    ans = (ans+tmp.m[0][6])%MOD;    cout << ans<<endl;}int main(){    int T;    scanf("%d",&T);    while(T--){        cin>> n>> aa >> bb;        if(n == 1)  cout << aa<<endl;        else if( n== 2) cout << bb<<endl;        else{            sov();        }    }}