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  • Google Code Jam 2008 Round 1A C Numbers(矩阵快速幂+化简方程,好题)

    Problem C. Numbers

    This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start Guide to get started.
    Small input
    15 points
     
    Large input
    35 points
     

    Problem

    In this problem, you have to find the last three digits before the decimal point for the number (3 + √5)n.

    For example, when n = 5, (3 + √5)5 = 3935.73982... The answer is 935.

    For n = 2, (3 + √5)2 = 27.4164079... The answer is 027.

    Input

    The first line of input gives the number of cases, TT test cases follow, each on a separate line. Each test case contains one positive integer n.

    Output

    For each input case, you should output:

    Case #X: Y
    where X is the number of the test case and Y is the last three integer digits of the number (3 + √5)n. In case that number has fewer than three integer digits, add leading zeros so that your output contains exactly three digits.

    Limits

    1 <= T <= 100

    Small dataset

    2 <= n <= 30

    Large dataset

    2 <= n <= 2000000000

    Sample


    Input 
     

    Output 
     
    2
    5
    2
    Case #1: 935
    Case #2: 027

    题目链接:Problem C. Numbers

    挑战编程书上的题目,跟HDU的4565有点像,只是这题a与b固定,要求的是整数部分的最后三位数字,不足补0。

    一开始书上的公式看不懂,问了下同学才弄懂,可以参考草稿纸上写的推出$a_n$的公式

    按照这个思路可以知道题目中所求的答案就是$2*a_n-1$了

    然后构造矩阵去求$a_n$即可

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define LC(x) (x<<1)
    #define RC(x) ((x<<1)+1)
    #define MID(x,y) ((x+y)>>1)
    #define CLR(arr,val) memset(arr,val,sizeof(arr))
    #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
    typedef pair<int, int> pii;
    typedef long long LL;
    const double PI = acos(-1.0);
    const int N = 2;
    int a = 3, b = 5, n, m = 1000;
    
    int mul(int a, int b)
    {
        int r = 0;
        while (b)
        {
            if (b & 1)
                r = (r + a) % m;
            a = (a << 1) % m;
            b >>= 1;
        }
        return r;
    }
    struct Mat
    {
        int A[N][N];
        void zero()
        {
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < N; ++j)
                    A[i][j] = 0;
        }
        void one()
        {
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < N; ++j)
                    A[i][j] = (i == j);
        }
        Mat operator*(Mat b)
        {
            Mat c;
            c.zero();
            for (int k = 0; k < N; ++k)
            {
                for (int i = 0; i < N; ++i)
                {
                    if (A[i][k])
                    {
                        for (int j = 0; j < N; ++j)
                        {
                            if (b.A[k][j])
                                c.A[i][j] = (c.A[i][j] + mul(A[i][k], b.A[k][j])) % m;
                        }
                    }
                }
            }
            return c;
        }
        friend Mat operator^(Mat a, int b)
        {
            Mat r;
            r.one();
            while (b)
            {
                if (b & 1)
                    r = r * a;
                a = a * a;
                b >>= 1;
            }
            return r;
        }
    };
    int main(void)
    {
        while (~scanf("%d", &n))
        {
            Mat A, B;
            A.zero();
            B.zero();
            A.A[0][0] = 1; A.A[0][1] = 0;
            A.A[1][0] = 1; A.A[1][1] = 0;
            B.A[0][0] = a; B.A[0][1] = 1;
            B.A[1][0] = b; B.A[1][1] = a;
            A = A * (B ^ n);
            printf("Case #%d: %03d
    ", q, ((A.A[0][0] << 1) - 1) % m);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Blackops/p/7107320.html
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