第一次用lrj的高精度类模板，感觉还是很好用的
c[x]表示数字x需要的火柴根数
将已经使用的火柴数i看做状态，每添加一个数字x状态就从i转移到i+c[x]
d[i]表示从节点0到节点i路径的条数，则答案f(n) = d(1) + d(2) + …… + d(n)
开始的时候不计入0，最后的时候如果n≥6答案再加一
1 #define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 struct bign 8 { 9 int len, s[1000]; 10 bign() { memset(s, 0, sizeof(s)); len = 1; } 11 bign operator = (const char* num) 12 { 13 len = strlen(num); 14 for(int i = 0; i < len; ++i) 15 s[i] = num[len - i - 1] - '0'; 16 return *this; 17 } 18 bign operator = (int num) 19 { 20 char s[1000]; 21 sprintf(s, "%d", num); 22 *this = s; 23 return *this; 24 } 25 bign (int num) { *this = num; } 26 bign (const char* num) { *this = num; } 27 string str() const 28 { 29 string res = ""; 30 for(int i = 0; i < len; ++i) 31 res = (char)(s[i] + '0') + res; 32 if(res == "") res = "0"; 33 return res; 34 } 35 36 bign operator + (const bign b) const 37 { 38 bign c; 39 c.len = 0; 40 for(int i = 0, g = 0; g || i < max(len, b.len); ++i) 41 { 42 int x = g; 43 if(i < len) x += s[i]; 44 if(i < b.len) x += b.s[i]; 45 c.s[c.len++] = x % 10; 46 g = x / 10; 47 } 48 return c; 49 } 50 51 bign operator += (const bign& b) 52 { 53 *this = *this + b; 54 return *this; 55 } 56 }; 57 58 istream& operator >> (istream& in, bign& x) 59 { 60 string s; 61 in >> s; 62 x = s.c_str(); 63 return in; 64 } 65 66 ostream& operator << (ostream &out, const bign& x) 67 { 68 out << x.str(); 69 return out; 70 } 71 72 const int maxn = 2000; 73 bign d[maxn + 10]; 74 int c[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6}; 75 76 int main(void) 77 { 78 #ifdef LOCAL 79 freopen("11375in.txt", "r", stdin); 80 #endif 81 82 memset(d, 0, sizeof(d)); 83 d[0] = 1; 84 for(int i = 0; i <= maxn; ++i) 85 for(int j = 0; j < 10; ++j) 86 if(!(i==0 && j==0) && i+c[j]<=maxn) 87 d[i+c[j]] += d[i]; 88 for(int i = 2; i <= maxn; ++i) 89 d[i] += d[i-1]; 90 int x; 91 while(scanf("%d", &x) == 1) 92 { 93 if(x < 6) cout << d[x] << endl; 94 else cout << d[x] + 1 << endl; 95 } 96 return 0; 97 }