题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=5950
题意:
已知递推公式: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)等于多少?
思路:
我xjb凑的,7*7的矩阵,主要是n^4怎么拆,这里有个很有道理的拆开n^4:http://blog.csdn.net/spring371327/article/details/52973534
代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define MS(a) memset(a,0,sizeof(a)) 5 #define MP make_pair 6 #define PB push_back 7 const int INF = 0x3f3f3f3f; 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 9 inline ll read(){ 10 ll x=0,f=1;char ch=getchar(); 11 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 13 return x*f; 14 } 15 ////////////////////////////////////////////////////////////////////////// 16 const int maxn = 1e5+10; 17 const ll mod = 2147493647; 18 19 ll N,A,B; 20 21 struct node{ 22 ll m[8][8]; 23 void clear(){ 24 MS(m); 25 } 26 }a; 27 28 node mul(node a,node b){ 29 node re; re.clear(); 30 for(int i=0; i<7; i++){ 31 for(int k=0; k<7; k++){ 32 if(a.m[i][k]){ 33 for(int j=0; j<7; j++) 34 re.m[i][j] = (re.m[i][j]+(a.m[i][k]*b.m[k][j])%mod)%mod; 35 } 36 } 37 } 38 return re; 39 } 40 41 node qpow(ll k){ 42 node re; 43 re.clear(); 44 for(int i=0; i<7; i++) re.m[i][i]=1; 45 while(k){ 46 if(k&1) re=mul(re,a); 47 a = mul(a,a); 48 k >>= 1; 49 } 50 return re; 51 } 52 53 int main(){ 54 int T = read(); 55 while(T--){ 56 a.clear(); 57 cin>>N>>A>>B; 58 a.m[0][0]=1,a.m[0][1]=2,a.m[0][2]=1,a.m[0][3]=8,a.m[0][4]=24,a.m[0][5]=32,a.m[0][6]=16; 59 a.m[1][0]=1,a.m[1][1]=0,a.m[1][2]=0,a.m[1][3]=0,a.m[1][4]=0,a.m[1][5]=0,a.m[1][6]=0; 60 a.m[2][0]=0,a.m[2][1]=0,a.m[2][2]=1,a.m[2][3]=4,a.m[2][4]=6,a.m[2][5]=4,a.m[2][6]=1; 61 a.m[3][0]=0,a.m[3][1]=0,a.m[3][2]=0,a.m[3][3]=1,a.m[3][4]=3,a.m[3][5]=3,a.m[3][6]=1; 62 a.m[4][0]=0,a.m[4][1]=0,a.m[4][2]=0,a.m[4][3]=0,a.m[4][4]=1,a.m[4][5]=2,a.m[4][6]=1; 63 a.m[5][0]=0,a.m[5][1]=0,a.m[5][2]=0,a.m[5][3]=0,a.m[5][4]=0,a.m[5][5]=1,a.m[5][6]=1; 64 a.m[6][0]=0,a.m[6][1]=0,a.m[6][2]=0,a.m[6][3]=0,a.m[6][4]=0,a.m[6][5]=0,a.m[6][6]=1; 65 ll r1=B, r2=A; 66 if(N==1) { cout<<r2<<endl; } 67 else if(N==2) { cout<<r1<<endl; } 68 else{ 69 a = qpow(N-2); 70 // cout << a.m[0][0] << " " << a.m[0][1] << " "<< a.m[0][2] << " " << a.m[0][3] << " "<< a.m[0][4] << " " << a.m[0][5] << " " << a.m[0][6] << endl; 71 ll ans = ((a.m[0][0]*B)%mod+(a.m[0][1]*A)%mod+a.m[0][2]+a.m[0][3]+a.m[0][4]+a.m[0][5]+a.m[0][6])%mod; 72 cout << ans << endl; 73 } 74 } 75 76 return 0; 77 }