UVa 10518 - How Many Calls?

传送门

UVa 10518 - How Many Calls?

题意

求第n项Fibonacci数列的计算次数。

思路

可以通过打表发现次数为2 * F(n) - 1。所以就变成了计算F(n)。

可是n很大,$O(n)$也会超时,然后就学习了矩阵快速幂。

代码

#include <bits/stdc++.h>
#define LL long long
using namespace std;
 
struct MATRIX
{
    LL mat[2][2];
};
 
MATRIX ans, temp;
LL mod;
 
MATRIX Calc(MATRIX a, MATRIX b)
{
    MATRIX t;
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
            t.mat[i][j] = (a.mat[i][0] * b.mat[0][j] + a.mat[i][1] * b.mat[1][j]) % mod;
    return t;
}
 
void Init()
{
    temp.mat[0][0] = 1, temp.mat[0][1] = 1;
    temp.mat[1][0] = 1, temp.mat[1][1] = 0;
    ans.mat[0][0] = ans.mat[1][1] = 1;
    ans.mat[0][1] = ans.mat[1][0] = 0;
}
 
int main()
{
    //freopen("input.txt", "r", stdin);
    int i, j, cases = 0;
    LL n;
    while (scanf("%lld%lld", &n, &mod), n + mod)
    {
        printf("Case %d: %lld %lld ", ++cases, n, mod);
        n++;
        Init();
        while (n)
        {
            if (n & 1)
                ans = Calc(temp, ans);
            temp = Calc(temp, temp);
            n >>= 1;
        }
        printf("%lld\n", (ans.mat[1][0] * 2 - 1 + mod) % mod);
    }
    return 0;
}

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