hdu 1250 Hat's Fibonacci

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1250

Hat's Fibonacci

Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
$F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n > 4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)$
Your task is to take a number as input, and print that Fibonacci number.

Input

Each line will contain an integers. Process to end of file.

Output

For each case, output the result in a line.

SampleInput

100

SampleOutput

4203968145672990846840663646

大数加法模板题。。

  1 #include<algorithm>

  2 #include<iostream>

  3 #include<cstdlib>

  4 #include<cstring>

  5 #include<cassert>

  6 #include<cstdio>

  7 #include<vector>

  8 #include<string>

  9 #include<map>

 10 #include<set>

 11 using std::cin;

 12 using std::max;

 13 using std::cout;

 14 using std::endl;

 15 using std::string;

 16 using std::istream;

 17 using std::ostream;

 18 #define sz(c) (int)(c).size()

 19 #define all(c) (c).begin(), (c).end()

 20 #define iter(c) decltype((c).begin())

 21 #define cls(arr,val) memset(arr,val,sizeof(arr))

 22 #define cpresent(c, e) (find(all(c), (e)) != (c).end())

 23 #define rep(i, n) for (int i = 0; i < (int)(n); i++)

 24 #define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)

 25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)

 26 #define pb(e) push_back(e)

 27 #define mp(a, b) make_pair(a, b)

 28 struct BigN {

 29     typedef unsigned long long ull;

 30     static const int Max_N = 2010;

 31     int len, data[Max_N];

 32     BigN() { memset(data, 0, sizeof(data)), len = 0; }

 33     BigN(const int num) {

 34         memset(data, 0, sizeof(data));

 35         *this = num;

 36     }

 37     BigN(const char *num) {

 38         memset(data, 0, sizeof(data));

 39         *this = num;

 40     }

 41     void clear() { len = 0, memset(data, 0, sizeof(data)); }

 42     BigN& clean(){ while (len > 1 && !data[len - 1]) len--;  return *this; }

 43     string str() const {

 44         string res = "";

 45         for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0');

 46         if (res == "") res = "0";

 47         res.reserve();

 48         return res;

 49     }

 50     BigN operator = (const int num) {

 51         int j = 0, i = num;

 52         do data[j++] = i % 10; while (i /= 10);

 53         len = j;

 54         return *this;

 55     }

 56     BigN operator = (const char *num) {

 57         len = strlen(num);

 58         for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0';

 59         return *this;

 60     }

 61     BigN operator + (const BigN &x) const {

 62         BigN res;

 63         int n = max(len, x.len) + 1;

 64         for (int i = 0, g = 0; i < n; i++) {

 65             int c = data[i] + x.data[i] + g;

 66             res.data[res.len++] = c % 10;

 67             g = c / 10;

 68         }

 69         return res.clean();

 70     }

 71     BigN operator * (const BigN &x) const {

 72         BigN res;

 73         int n = x.len;

 74         res.len = n + len;

 75         for (int i = 0; i < len; i++) {

 76             for (int j = 0, g = 0; j < n; j++) {

 77                 res.data[i + j] += data[i] * x.data[j];

 78             }

 79         }

 80         for (int i = 0; i < res.len - 1; i++) {

 81             res.data[i + 1] += res.data[i] / 10;

 82             res.data[i] %= 10;

 83         }

 84         return res.clean();

 85     }

 86     BigN operator * (const int num) const {

 87         BigN res;

 88         res.len = len + 1;

 89         for (int i = 0, g = 0; i < len; i++) res.data[i] *= num;

 90         for (int i = 0; i < res.len - 1; i++) {

 91             res.data[i + 1] += res.data[i] / 10;

 92             res.data[i] %= 10;

 93         }

 94         return res.clean();

 95     }

 96     BigN operator - (const BigN &x) const {

 97         assert(x <= *this);

 98         BigN res;

 99         for (int i = 0, g = 0; i < len; i++) {

100             int c = data[i] - g;

101             if (i < x.len) c -= x.data[i];

102             if (c >= 0) g = 0;

103             else g = 1, c += 10;

104             res.data[res.len++] = c;

105         }

106         return res.clean();

107     }

108     BigN operator / (const BigN &x) const {

109         BigN res, f = 0;

110         for (int i = len - 1; ~i; i--) {

111             f *= 10;

112             f.data[0] = data[i];

113             while (f >= x) {

114                 f -= x;

115                 res.data[i]++;

116             }

117         }

118         res.len = len;

119         return res.clean();

120     }

121     BigN operator % (const BigN &x) {

122         BigN res = *this / x;

123         res = *this - res * x;

124         return res;

125     }

126     BigN operator += (const BigN &x) { return *this = *this + x; }

127     BigN operator *= (const BigN &x) { return *this = *this * x; }

128     BigN operator -= (const BigN &x) { return *this = *this - x; }

129     BigN operator /= (const BigN &x) { return *this = *this / x; }

130     BigN operator %= (const BigN &x) { return *this = *this % x; }

131     bool operator <  (const BigN &x) const {

132         if (len != x.len) return len < x.len;

133         for (int i = len - 1; ~i; i--) {

134             if (data[i] != x.data[i]) return data[i] < x.data[i];

135         }

136         return false;

137     }

138     bool operator >(const BigN &x) const { return x < *this; }

139     bool operator<=(const BigN &x) const { return !(x < *this); }

140     bool operator>=(const BigN &x) const { return !(*this < x); }

141     bool operator!=(const BigN &x) const { return x < *this || *this < x; }

142     bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }

143     friend istream& operator >> (istream &in, BigN &x) {

144         string src;

145         in >> src;

146         x = src.c_str();

147         return in;

148     }

149     friend ostream& operator << (ostream &out, const BigN &x) {

150         out << x.str();

151         return out;

152     }

153 }A[5];

154 void solve(int n) {

155     fork(i, 1, 4) A[i] = 1;

156     if (n < 5) cout << A[n] << endl;

157     else {

158         int x = 1;

159         fork(i, 5, n) {

160             A[0] = A[1] + A[2] + A[3] + A[4];

161             A[x] = A[0];

162             if (++x == 5) x = 1;

163         }

164         cout << A[0] << endl;

165     }

166     rep(i, 5) A[i].clear();

167 }

168 int main() {

169 #ifdef LOCAL

170     freopen("in.txt", "r", stdin);

171     freopen("out.txt", "w+", stdout);

172 #endif

173     std::ios::sync_with_stdio(false);

174     int n;

175     while (cin >> n) {

176         solve(n);

177     }

178     return 0;

179 }

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