P1149

　　这题不难，我写的一个复杂度 $ O(n^2) $ 的递归算法。。

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i < b; i++)
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define index(a) (a - 'A')
#define transUpp(a) (a - 32)
#define transLow(a) (a + 32)
#define ll long long
#define PB push_back
int gcd(int a, int b){return b == 0 ? a : gcd(a%b, a);}
const int N = 782;
int num[N] = { 6,2,5,5,4,5,6,3,7,6 }, has[N][N];
int counts = 0;
void re(int n, int l1, int l2)
{
    if (l1 + l2 < N)
    {
        has[l1][l2] = 1;
        if (n == num[l1] + num[l2] + num[l1 + l2]) counts++;
        if (!has[l1][l2 + 1]) re(n, l1, l2 + 1);
        if (!has[l1 + 1][l2]) re(n, l1 + 1, l2);
    }
}
int main()
{
    // rep(i, 0, N) rep(j, 0, N) has[i][j] = 0; // 去掉这行初始化,投机优化一点速度
    for (int k = 10; k < N; k++)
        num[k] = num[k%10] + num[k/10];
    int n;
    cin >> n;
    re(n - 4, 0, 0);
    cout << counts << endl;
    return 0;
}