HDU5950 Recursive sequence —— 矩阵快速幂
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5950
Recursive sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1345 Accepted Submission(s): 606
Problem Description
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 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.
Input
The 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.
Each case contains only one line with three numbers N, a and b where N,a,b <
Output
For 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 Input
23 1 24 1 10
Sample Output
85369HintIn 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.
题解:
典型的矩阵快速幂的运用。关键是i^4怎么维护?我们可以当成求第i+1项,那么i^4就变成了(i+1)^4。那么这时我们可以用二项式定理从i^4、i^3、i^2、i^1、i^0的组合中得到(i+1)^4。也就是说总共需要维护:f[i+1]、f[i]、(i+1)^4、(i+1)^3、(i+1)^2、(i+1)^1、(i+1)^0。矩阵如下:
代码如下:
#include <bits/stdc++.h>#define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)#define ms(a,b) memset((a),(b),sizeof((a)))using namespace std;typedef long long LL;const LL mod = 2147493647;const int maxn = 1e5;struct Mat{ LL mat[7][7]; void init() { rep(i,0,6) rep(j,0,6) mat[i][j] = (i==j); }};Mat p = { 1, 2, 1, 4, 6, 4, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0 ,0, 0, 1 };Mat mul(Mat x, Mat y){ Mat s; ms(s.mat,0); rep(i,0,6) rep(j,0,6) rep(k,0,6) s.mat[i][j] += (x.mat[i][k]*y.mat[k][j])%mod, s.mat[i][j] %= mod; return s;}Mat qpow(Mat x, LL y){ Mat s; s.init(); while(y) { if(y&1) s = mul(s, x); x = mul(x, x); y >>= 1; } return s;}int main(){ int T; scanf("%d",&T); while(T--) { LL n, a, b; scanf("%lld%lld%lld",&n,&a,&b); if(n == 1) { printf("%lld\n",a); continue; } if(n == 2) { printf("%lld\n",b); continue; } Mat x = p; x = qpow(x, n-2); LL ans = 0; ans = (ans + b*x.mat[0][0]) % mod; ans = (ans + a*x.mat[0][1]%mod) % mod; ans = (ans + 16*x.mat[0][2]%mod) % mod; ans = (ans + 8*x.mat[0][3]%mod) % mod; ans = (ans + 4*x.mat[0][4]%mod) % mod; ans = (ans + 2*x.mat[0][5]%mod) % mod; ans = (ans+x.mat[0][6]) % mod; printf("%lld\n",ans); }}
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