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):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }