Binary Strings Gym - 101161G 矩阵快速幂 + 打表

http://codeforces.com/gym/101161/attachments

这题通过打表，可以知道长度是i的时候的合法方案数。

然后得到f[1] = 2, f[2] = 3, f[3] = 5, f[4] = 8......这样的广义fib数列

现在要求f[k] + f[2k] + f[3k] + ...... + f[xk]的总和。

直接做很难做，我不知道f[i * k] = x * f[(i - 1) * k] + y * f[(i - 2) * k]

推不出系数的话，有一个结论就是：fib的任意一项肯定能表示成x * f(i - 1) + y * f(i - 2)，就是两个连续的fib数字，能表示出后面所有的fib数列。知道这样的东西后。

可以知道，比如k = 2

形如f(6) = 2 * f(4) + f(3)、然后f(8) = 2 * f(6) + f(5)

那么就可以构造矩阵了。

sum, f(4)，f(3)   ----->   newSum, f(6), f(5)

同样用f(4)和f(3)把f(5)表示出来即可（注意：肯定能表示）

 

一般构造了矩阵后，写一个求第k项的函数出来，往往比较有用

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#define X first
#define Y second
#define lson T[rt].l
#define rson T[rt].r
#define clr(u,v); memset(u,v,sizeof(u));
#define in() freopen("data.txt","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back

using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int maxn = 5;
const int MOD = 1e9 + 7;
struct Matrix {
    LL a[maxn][maxn];
    int row, col;
} base, a;
struct Matrix mul(struct Matrix a, struct Matrix b, int MOD) {
    struct Matrix c = {0};
    c.row = a.row, c.col = b.col;
    for (int i = 1; i <= a.row; ++i) {
        for (int j = 1; j <= b.col; j++) {
            for (int k = 1; k <= b.row; ++k) {
                c.a[i][j] += a.a[i][k] * b.a[k][j];
                c.a[i][j] %= MOD;
            }
        }
    }
    return c;
}
struct Matrix qp(Matrix ans, Matrix base, LL n, LL MOD) {
    while (n) {
        if (n & 1) {
            ans = mul(ans, base, MOD);
        }
        n >>= 1;
        base = mul(base, base, MOD);
    }
    return ans;
}
int f;
void initBaseFib() {
    base.row = base.col = 2;
    base.a[1][1] = 1, base.a[1][2] = 1;
    base.a[2][1] = 1, base.a[2][2] = 0;
}
LL findK(int a, int b, LL k) {
    if (k == 1) return a;
    if (k == 2) return b;
    initBaseFib();
    Matrix t;
    t.row = 1, t.col = 2;
    t.a[1][1] = b, t.a[1][2] = a;
    t = qp(t, base, k - 2, MOD);
    return t.a[1][1];
}
void initBaseSum(int a, int b, int c, int d) {
    base.row = base.col = 3;
    base.a[1][1] = 1, base.a[1][2] = 0, base.a[1][3] = 0;
    base.a[2][1] = 1, base.a[2][2] = a, base.a[2][3] = b;
    base.a[3][1] = 0, base.a[3][2] = c, base.a[3][3] = d;
}
int pre[22];
int sum[22];
LL getpos(LL n, LL k) {
    if (n == 0) return 0;
    if (n == 1) return findK(2, 3, n * k);
    if (k == 2) {
        initBaseSum(2, 1, 1, 1);
        Matrix t;
        t.row = 1, t.col = 3;
        t.a[1][1] = 3, t.a[1][2] = 8, t.a[1][3] = 5;
        t = qp(t, base, n - 1, MOD);
        return t.a[1][1];
    } else if (k == 3) {
        initBaseSum(3, 2, 2, 1);
        Matrix t;
        t.row = 1, t.col = 3;
        t.a[1][1] = 5, t.a[1][2] = 21, t.a[1][3] = 13;
        t = qp(t, base, n - 1, MOD);
        return t.a[1][1];
    } else {
        Matrix t;
        t.row = 1, t.col = 3;
        t.a[1][1] = findK(2, 3, k), t.a[1][2] = findK(2, 3, 2 * k), t.a[1][3] = findK(2, 3, 2 * k - 1);
        initBaseSum(findK(2,3,k-1), findK(2,3,k-2), findK(2,3,k-2), findK(2,3,k-3));
        t = qp(t, base, n - 1, MOD);
        return t.a[1][1];
    }
}
void work() {
    LL be, en, k;
    cin >> be >> en >> k;
    if (k == 1) {
        //sumofall fib
        if (en <= 2) {
            printf("Case %d: %d\n", ++f, sum[en] - sum[be - 1]);
            return;
        }

        initBaseSum(1, 1, 1, 0);
        be--;
        a.row = 1, a.col = 3;
        a.a[1][1] = 2, a.a[1][2] = 3, a.a[1][3] = 2;
        Matrix ret = qp(a, base, en - 1, MOD);
        LL resEn = ret.a[1][1];
        LL resBe = 0;
        if (be - 1 >= 0) {
            ret = qp(a, base, be - 1, MOD);
            resBe = ret.a[1][1];
        }
        printf("Case %d: %d\n", ++f, (resEn - resBe + MOD) % MOD);
        return;
    } else {
        en = en / k;
        if (be % k == 0) be = (be - k) / k;
        else be /= k;
//        cout << be << " " << en << endl;
        LL ans = (getpos(en, k) - getpos(be, k) + MOD) % MOD;
        printf("Case %d: %d\n", ++f, ans);
    }
}
int cnt[22];
void dfs(int k, int t1, int t2) {
    if (k == t1 || k == t2) {
        cnt[k]++;
        return;
    }
    dfs(k - 1, t1, t2);
    dfs(k - 2, t1, t2);
}
int main() {
#ifdef local
    in();
#else
#endif
    pre[1] = 2, pre[2] = 3, pre[3] = 5, pre[4] = 8;
    sum[1] = 2, sum[2] = 5, sum[3] = 10, sum[4] = 18;
//    int one = 6, t1 = 4, t2 = 2;
//    dfs(one, t1, t2);
//    cout << cnt[t1] << " " << cnt[t2] << endl;
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}