UVA10518 How Many Calls? —— 矩阵快速幂

题目链接：https://vjudge.net/problem/UVA-10518

题解：

问：求斐波那契数f[n]的时候调用了多少次f[n] = f[n-1] + f[n-2]，没有记忆化，一直递归到f[0]、f[1]，其中f[0]、f[1]也算调用了一次。

设求f[n]调用了S[n]次，则可知： S[n] = S[n-1] + S[n-2] + 1。构造矩阵求解即可。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 //const int MOD = 1e9+7;
 const int MAXN = 1e6+; int MOD;
 const int Size = ;
 struct MA
 {
     LL mat[Size][Size];
     void init()
     {
         for(int i = ; i<Size; i++)
         for(int j = ; j<Size; j++)
             mat[i][j] = (i==j);
     }
 }; MA mul(MA x, MA y)
 {
     MA ret;
     memset(ret.mat, , sizeof(ret.mat));
     for(int i = ; i<Size; i++)
     for(int j = ; j<Size; j++)
     for(int k = ; k<Size; k++)
         ret.mat[i][j] += 1LL*x.mat[i][k]*y.mat[k][j]%MOD, ret.mat[i][j] %= MOD;;
     return ret;
 } MA qpow(MA x, LL y)
 {
     MA s;
     s.init();
     while(y)
     {
         if(y&) s = mul(s, x);
         x = mul(x, x);
         y >>= ;
     }
     return s;
 } MA tmp ={
     , , ,
     , , ,
     , ,
 }; int main()
 {
     LL n, b, f[] = {,}, kase = ;
     while(scanf("%lld%lld",&n,&b)&&(n||b))
     {
         MOD = b;
         if(n<=)
         {
             printf("Case %lld: %lld %lld %lld\n", ++kase, n, b, f[n]%MOD);
             continue;
         }         MA s = tmp;
         s = qpow(s, n-);
         LL ans = ((s.mat[][]+s.mat[][])%MOD+s.mat[][])%MOD;
         printf("Case %lld: %lld %lld %lld\n", ++kase, n, b, ans);
     }
 }