Recursive sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 249 Accepted Submission(s): 140Problem DescriptionFarmer 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 i4. 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.InputThe first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.OutputFor each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.Sample Input2 3 1 2 4 1 10Sample Output85 369HintIn the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.SourceRecommendjiangzijing2015 | We have carefully selected several similar problems for you: 5960 5959 5958 5957 5956Statistic | Submit | Discuss | Note
题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5950
题目大意：
Fi=Fi-1+2Fi-2+i4。给定F1和F2求Fn。
题目思路：
【递推+矩阵快速幂】
现场用算了1个多小时的公式过了。
主要还是我太菜。递推写的太少。
先考虑f(i)=f(i-1)+2f(i-2)，很容易写出递推矩阵
0 2
1 1
(i+1)4=i4+4i3+6i2+4i+1。
所以需要在递推矩阵种存下i的4 3 2 1 0次幂，以便推出(i+1)4，矩阵为
1 0 0 0 0
4 1 0 0 0
6 3 1 0 0
4 3 2 1 0
1 1 1 1 1
于是f={fi-1,fi,i4,i3,i2,i1,i0}，将以上两个矩阵合并，即可推出{fi,fi+1,(i+1)4,(i+1)3,(i+1)2,(i+1)1,(i+1)0}.矩阵如下
0 2 0 0 0 0 0
1 1 0 0 0 0 0
0 1 1 0 0 0 0
0 4 4 1 0 0 0
0 6 6 3 1 0 0
0 4 4 3 2 1 0
0 1 1 1 1 1 1
推出转移矩阵后只需要根据n求矩阵快速幂即可。
1 // 2 //by coolxxx 3 //#include<bits/stdc++.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #pragma comment(linker,"/STACK:1024000000,1024000000") 21 #define min(a,b) ((a)<(b)?(a):(b)) 22 #define max(a,b) ((a)>(b)?(a):(b)) 23 #define abs(a) ((a)>0?(a):(-(a))) 24 #define lowbit(a) (a&(-a)) 25 #define sqr(a) ((a)*(a)) 26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define eps (1e-8) 29 #define J 10000 30 #define mod 2147493647 31 #define MAX 0x7f7f7f7f 32 #define PI 3.14159265358979323 33 #define N 14 34 #define M 7 35 using namespace std; 36 typedef long long LL; 37 double anss; 38 LL aans; 39 int cas,cass; 40 int n,m,lll,ans; 41 LL f[N]; 42 LL a[N][N]; 43 LL ma[N][N]={{0}, 44 {0,0,2,0,0,0,0,0}, 45 {0,1,1,0,0,0,0,0}, 46 {0,0,1,1,0,0,0,0}, 47 {0,0,4,4,1,0,0,0}, 48 {0,0,6,6,3,1,0,0}, 49 {0,0,4,4,3,2,1,0}, 50 {0,0,1,1,1,1,1,1}}; 51 void multi(LL a[][N],LL b[][N],LL c[][N]) 52 { 53 int i,j,k; 54 LL t[N][N]; 55 mem(t,0); 56 for(i=1;i<=M;i++) 57 for(j=1;j<=M;j++) 58 for(k=1;k<=M;k++) 59 t[i][j]=(t[i][j]+a[i][k]*b[k][j]%mod)%mod; 60 memcpy(c,t,sizeof(t)); 61 } 62 void mi(LL a[][N],int y) 63 { 64 LL tmp[N][N]; 65 mem(tmp,0); 66 tmp[1][1]=tmp[2][2]=tmp[3][3]=tmp[4][4]=tmp[5][5]=tmp[6][6]=tmp[7][7]=1; 67 while(y) 68 { 69 if(y&1)multi(tmp,a,tmp); 70 y>>=1;multi(a,a,a); 71 } 72 memcpy(a,tmp,sizeof(tmp)); 73 } 74 void work() 75 { 76 LL t[N]; 77 mem(t,0); 78 int i,j; 79 for(i=1;i<=M;i++) 80 for(j=1;j<=M;j++) 81 t[i]=(t[i]+f[j]*a[j][i]%mod)%mod; 82 memcpy(f,t,sizeof(t)); 83 } 84 int main() 85 { 86 #ifndef ONLINE_JUDGE 87 freopen("1.txt","r",stdin); 88 // freopen("2.txt","w",stdout); 89 #endif 90 int i,j,k; 91 int x,y,z; 92 // init(); 93 for(scanf("%d",&cass);cass;cass--) 94 // for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 95 // while(~scanf("%s",s)) 96 // while(~scanf("%d%d",&n,&m)) 97 { 98 memcpy(a,ma,sizeof(a)); 99 scanf("%d%lld%lld",&n,&f[1],&f[2]); 100 f[3]=16,f[4]=8,f[5]=4,f[6]=2,f[7]=1; 101 if(n==1) 102 { 103 printf("%lld ",f[1]); 104 continue; 105 } 106 mi(a,n-2); 107 work(); 108 printf("%lld ",f[2]); 109 } 110 return 0; 111 } 112 /* 113 // 114 115 // 116 */