Gym 101161 G - Binary Strings[矩阵快速幂][dp]

题意：问长度在 [L,R] 之间的，且长度被 K 整除的，且仅包含 0 和 1 的，且不包含连续 1 的字符串的个数。其中 1≤L,R≤1018,3≤K≤109
分析：考虑 dp ， dp[i][0/1] 表示长度为 i ，最后一位字符是 0,1 的字符串个数。
转移： dp[i][0]=dp[i−1][0]+dp[i−1][1],dp[i][1]=dp[i−1][0]
再用矩阵加速 dp
A=⎛⎝⎜110100001⎞⎠⎟,B=⎛⎝⎜100010111⎞⎠⎟
那么 ansR=⎛⎝⎜100000000⎞⎠⎟∗(AK∗B)R/K
再减去 L−1 的数量就是答案。

#include
using namespace std;
#define ll long long
#define ull unsigned long long
#define lson l,mid,id<<1
#define rson mid+1,r,id<<1|1
#define lll __int128;
typedef pair<int, int> pii;
typedef pair pll;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 50010;
const int MAXM = 100005;
const ll MOD = 1000000007;
const double eps = 1e-8;
struct Mat {
    ll edge[3][3];
    friend Mat operator*(const Mat &a, const Mat &b) {
        Mat c;
        memset(c.edge, 0, sizeof(c.edge));
        for (int i = 0; i<3; ++i) {
            for (int j = 0; j<3; ++j) {
                for (int k = 0; k<3; ++k) {
                    c.edge[i][j] += a.edge[i][k] * b.edge[k][j];
                    c.edge[i][j] %= MOD;
                }
            }
        }
        return c;
    }
    void print() {
        for (int i = 0; i<3; ++i) {
            for (int j = 0; j<3; ++j) {
                printf("%I64d ", edge[i][j]);
            }
            printf("\n");
        }
    }
};

Mat qpow(Mat a, ll x) {
    Mat res;
    memset(res.edge, 0, sizeof(res.edge));
    for (int i = 0; i<3; ++i)res.edge[i][i] = 1;
    while (x) {
        if (x & 1)res = res * a;
        a = a * a;
        x >>= 1;
    }
    return res;
}
int main() {
    int T, cas = 0; scanf("%d", &T);
    ll L, R, K;
    while (T--) {
        scanf("%I64d%I64d%I64d", &L, &R, &K);
        Mat E;
        memset(E.edge, 0, sizeof(E.edge));
        E.edge[0][0] = 1;
        Mat F, P;
        memset(F.edge, 0, sizeof(F.edge));
        F.edge[0][0] = F.edge[0][1] = F.edge[1][0] = 1;
        F.edge[2][2] = 1;
        F = qpow(F, K);
        memset(P.edge, 0, sizeof(P.edge));
        P.edge[0][2] = P.edge[1][2] = P.edge[2][2] = 1;
        P.edge[0][0] = P.edge[1][1] = 1;
        F = F * P;
        //F.print();
        Mat a = E * qpow(F, (L - 1) / K);
        Mat b = E * qpow(F, R / K);
        // b.print();
        printf("Case %d: %I64d\n", ++cas, (b.edge[0][2] - a.edge[0][2] + MOD) % MOD);
    }
}