UVa 11375 Matches

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 }

代码君