Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo（矩阵快速幂）

题意：主角可以从 (0, 0)走到(x + 1, y + 1), (x + 1, y), 或者 (x + 1, y - 1)。有k个区间完全覆盖0到k，每个区间有一个值c[i]，当x在一个区间内时，主角不能走到大于c[i]或者小于0的位置。问走到 (k, 0)有多少种走法，答案mod1e9+7。

这题c[i]的范围很小，只有16。裸的做法就是dp，这个可以用矩阵快速幂来优化。

现场想了半天组合数，看来我还是Naive。赛后再做还挂了一发快速幂写错（日常），挂了一发longlong（还是日常）。

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3fLL
typedef long long ll;
const ll mod = 1e9 + 7;
#define int long long
int a[105], b[105], c[105];
struct Martix{
    ll a[20][20];
    void print(int n){
        for (int i = 0; i < n; i++){
            for (int j = 0; j < n; j++) cout << a[i][j] << ' ';
            cout << endl;
        }
    }
};
Martix mul(Martix m1, Martix m2, int& n){
    Martix ret;
    memset(ret.a, 0, sizeof(ret.a));
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            for (int k = 0; k < n; k++){
                ret.a[i][j] += m1.a[i][k] * m2.a[k][j] + mod;
                ret.a[i][j] %= mod;
            }
        }
    }
    return ret;
}
Martix pow(Martix m, int p, int n){
    Martix ret;
    memset(ret.a, 0, sizeof(ret.a));
    for (int i = 0; i < n; i++) ret.a[i][i] = 1;
    while (p){
        if (p & 1) ret = mul(ret, m, n);
        m = mul(m, m, n);
        p >>= 1;
    }
    return ret;
}
vector<int> mul(Martix m, vector<int> k, int n){
    vector<int>ret;
    ret.resize(20);
    for (int i = 0; i < n; i++){
        ret[i] = 0;
        for (int j = 0; j < n; j++){
            ret[i] += m.a[i][j] * k[j] + mod;
            ret[i] %= mod;
        }
    }
    for (int i = n; i < 20; i++) ret[i] = 0;
    return ret;
}
Martix getMartix(int n){
    Martix ret;
    memset(ret.a, 0, sizeof(ret.a));
    for (int i = 0; i < n; i++){
        for (int j = max(0LL, i - 1); j <= min(n, i + 1); j++){
            ret.a[i][j] = 1;
        }
    }
    return ret;
}
vector<int> solve(vector<int> ans, int i){
    Martix m = getMartix(c[i] + 1);
    //m.print(c[i] + 1);
    m = pow(m, b[i] - a[i], c[i] + 1);
    //m.print(c[i] + 1);
    return mul(m, ans, c[i] + 1);
}
main(){
    //ios::sync_with_stdio(0);
    int n, k; cin >> n >> k;
    for (int i = 0; i < n; i++){
        cin >> a[i] >> b[i] >> c[i];
    }
    b[n - 1] = k;
    vector<int>ans; ans.resize(20);
    ans[0] = 1;
    for (int i = 0; i < n; i++){
        ans = solve(ans, i);
    }
    cout << ans[0] << endl;
    return 0;
}