HDU 1042 N!（高精度乘）

Problem Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

 

Input

One N in one line, process to the end of file.

 

Output

For each N, output N! in one line.

 

题目大意：求N的阶乘。

思路：用高精度，内存存不下这么多只能每次都重新算了……

 

代码（3093MS）：

  1 //模板测试

  2 #include <iostream>

  3 #include <cstdio>

  4 #include <cstring>

  5 #include <string>

  6 #include <algorithm>

  7 using namespace std;

  8 

  9 const int MAXN = 100010;

 10 

 11 struct bign {

 12     int len, s[MAXN];

 13 

 14     bign () {

 15         memset(s, 0, sizeof(s));

 16         len = 1;

 17     }

 18     bign (int num) { *this = num; }

 19     bign (const char *num) { *this = num; }

 20 

 21     void clear() {

 22         memset(s, 0, sizeof(s));

 23         len = 1;

 24     }

 25 

 26     bign operator = (const int num) {//数字

 27         char s[MAXN];

 28         sprintf(s, "%d", num);

 29         *this = s;

 30         return *this;

 31     }

 32     bign operator = (const char *num) {//字符串

 33         for(int i = 0; num[i] == '0'; num++) ;  //去前导0

 34         if(*num == 0) --num;

 35         len = strlen(num);

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

 37         return *this;

 38     }

 39 

 40     bign operator + (const bign &b) const {

 41         bign c;

 42         c.len = 0;

 43         for(int i = 0, g = 0; g || i < max(len, b.len); ++i) {

 44             int x = g;

 45             if(i < len) x += s[i];

 46             if(i < b.len) x += b.s[i];

 47             c.s[c.len++] = x % 10;

 48             g = x / 10;

 49         }

 50         return c;

 51     }

 52 

 53     bign operator += (const bign &b) {

 54         *this = *this + b;

 55         return *this;

 56     }

 57 

 58     void clean() {

 59         while(len > 1 && !s[len-1]) len--;

 60     }

 61 

 62     bign operator * (const bign &b) {

 63         bign c;

 64         c.len = len + b.len;

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

 66             for(int j = 0; j < b.len; ++j) {

 67                 c.s[i+j] += s[i] * b.s[j];

 68             }

 69         }

 70         for(int i = 0; i < c.len; ++i) {

 71             c.s[i+1] += c.s[i]/10;

 72             c.s[i] %= 10;



 73         }

 74         c.clean();

 75         return c;

 76     }

 77     bign operator *= (const bign &b) {

 78         *this = *this * b;

 79         return *this;

 80     }

 81 

 82     bign operator *= (const int &b) {//使用前要保证>len的位置都是空的

 83         for(int i = 0; i < len; ++i) s[i] *= b;

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

 85             s[i + 1] += s[i] / 10;

 86             s[i] %= 10;

 87         }

 88         while(s[len]) {

 89             s[len + 1] += s[len] / 10;

 90             s[len] %= 10;

 91             ++len;

 92         }

 93         return *this;

 94     }

 95 

 96     bign operator - (const bign &b) {

 97         bign c;

 98         c.len = 0;

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

100             int x = s[i] - g;

101             if(i < b.len) x -= b.s[i];

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

103             else {

104                 g = 1;

105                 x += 10;

106             }

107             c.s[c.len++] = x;

108         }

109         c.clean();

110         return c;

111     }

112     bign operator -= (const bign &b) {

113         *this = *this - b;

114         return *this;

115     }

116 

117     bign operator / (const bign &b) {

118         bign c, f = 0;

119         for(int i = len - 1; i >= 0; i--) {

120             f *= 10;

121             f.s[0] = s[i];

122             while(f >= b) {

123                 f -= b;

124                 c.s[i]++;

125             }

126         }

127         c.len = len;

128         c.clean();

129         return c;

130     }

131     bign operator /= (const bign &b) {

132         *this  = *this / b;

133         return *this;

134     }

135 

136     bign operator % (const bign &b) {

137         bign r = *this / b;

138         r = *this - r*b;

139         return r;

140     }

141     bign operator %= (const bign &b) {

142         *this = *this % b;

143         return *this;

144     }

145 

146     bool operator < (const bign &b) {

147         if(len != b.len) return len < b.len;

148         for(int i = len-1; i >= 0; i--) {

149             if(s[i] != b.s[i]) return s[i] < b.s[i];

150         }

151         return false;

152     }

153 

154     bool operator > (const bign &b) {

155         if(len != b.len) return len > b.len;

156         for(int i = len-1; i >= 0; i--) {

157             if(s[i] != b.s[i]) return s[i] > b.s[i];

158         }

159         return false;

160     }

161 

162     bool operator == (const bign &b) {

163         return !(*this > b) && !(*this < b);

164     }

165 

166     bool operator != (const bign &b) {

167         return !(*this == b);

168     }

169 

170     bool operator <= (const bign &b) {

171         return *this < b || *this == b;

172     }

173 

174     bool operator >= (const bign &b) {

175         return *this > b || *this == b;

176     }

177 

178     string str() const {

179         string res = "";

180         for(int i = 0; i < len; ++i) res = char(s[i]+'0') + res;

181         return res;

182     }

183 };

184 

185 istream& operator >> (istream &in, bign &x) {

186     string s;

187     in >> s;

188     x = s.c_str();

189     return in;

190 }

191 

192 ostream& operator << (ostream &out, const bign &x) {

193     out << x.str();

194     return out;

195 }

196 

197 bign ans;

198 

199 void solve(int n) {

200     ans.clear();

201     ans.len = ans.s[0] = 1;

202     for(int i = 2; i <= n; ++i) ans *= i;

203     cout<<ans<<endl;

204 }

205 

206 int main() {

207     int n;

208     while(scanf("%d", &n)!=EOF) {

209         //cout<<f[n]<<endl;

210         solve(n);

211     }

212     return 0;

213 }

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