[POJ 3734] Blocks (矩阵快速幂、组合数学）

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define For(i, n) for (int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
const int maxn = 5;
const int maxm = 5;
const int mod = 10007;
struct Matrix {
    int n, m;
    ll a[maxn][maxm];
    void clear() {
        n = m = 0;
        memset(a, 0, sizeof(a));
    }
    Matrix operator * (const Matrix &b) const {
        Matrix tmp;
        tmp.clear();
        tmp.n = n; tmp.m = b.m;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                if (!a[i][j]) continue;  //稀疏矩阵乘法优化
                for (int k = 0; k < b.m; k++) {
                    tmp.a[i][k] += a[i][j] * b.a[j][k];
                    tmp.a[i][k] %= mod;
                }
            }
        return tmp;
    }
};
int n;
Matrix Matrix_pow(Matrix A, int k) {
    Matrix res;
    res.clear();
    res.n = res.m = 4;
    for (int i = 0; i < 4; i++) res.a[i][i] = 1;
    while(k) {
        if (k & 1) res = res * A;
        k >>= 1;
        A = A * A;
    }
    return res;
}
int main () {
    int t;
    scanf("%d", &t);
    Matrix A;
    A.clear();
    A.n = A.m = 4;
    A.a[0][0] = 2; A.a[0][1] = 1; A.a[0][2] = 1; A.a[0][3] = 0;
    A.a[1][0] = 1; A.a[1][1] = 2; A.a[1][2] = 0; A.a[1][3] = 1;
    A.a[2][0] = 1; A.a[2][1] = 0; A.a[2][2] = 2; A.a[2][3] = 1;
    A.a[3][0] = 0; A.a[3][1] = 1; A.a[3][2] = 1; A.a[3][3] = 2;
    while(t--) {
        scanf("%d", &n);
        Matrix res = Matrix_pow(A, n);
        printf("%d\n", res.a[0][0]);
    }
    return 0;
}

/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define For(i, n) for (int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
const int mod = 10007;
int multi_pow(int a, int k, int mod) {
    int ans = 1;
    while(k) {
        if (k & 1) ans = (ans * a) % mod;
        k >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}
int main () {
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int ans = multi_pow(2, n - 1, mod);
        ans = (ans * ans) % mod + ans;
        printf("%d\n", ans % mod);
    }
    return 0;
}