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

Problem C. Numbers

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Small input 15 points	Solve C-small
Large input 35 points	Solve C-large

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, T. T 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\n", q, ((A.A[0][0] << 1) - 1) % m);
    }
    return 0;
}