nyoj 28 大数阶乘
题目链接：nyoj 28
就是个简单的高精度，只是一开始我打表超内存了，然后用了各种技巧硬是把内存缩到了题目要求以下（5w+kb），感觉挺爽的，代码如下：
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 using namespace std; 6 typedef long long LL; 7 const int M = 5002; 8 const int mod = 1e6; 9 10 int b[M + 3][2800], len[M + 3]; 11 12 inline void init(int n = M) { 13 b[1][0] = 1; len[1] = 1; 14 for(int i = 2; i <= n; ++i) { 15 int x = len[i - 1]; 16 int carry = 0; 17 LL tmp; 18 for(int j = 0; j < x; ++j) { 19 tmp = carry + (LL)b[i - 1][j] * i; 20 b[i][j] = tmp % mod; 21 carry = tmp / mod; 22 } 23 while(carry) { 24 b[i][x++] = carry % mod; 25 carry /= mod; 26 } 27 len[i] = x; 28 } 29 } 30 31 inline void print(const int &x) { 32 putchar(x / 100000 % 10 + '0'); 33 putchar(x / 10000 % 10 + '0'); 34 putchar(x / 1000 % 10 + '0'); 35 putchar(x / 100 % 10 + '0'); 36 putchar(x / 10 % 10 + '0'); 37 putchar(x % 10 + '0'); 38 } 39 40 inline void output(const int &n) { 41 int i = len[n] - 1; 42 const int &x = b[n][i]; 43 printf("%d",x); 44 45 for(--i; i >= 0; --i) 46 print(b[n][i]); 47 puts(""); 48 } 49 50 int main() { 51 int m; 52 init(); 53 while(~scanf("%d",&m)) 54 output(m); 55 return 0; 56 }
出题人原意应该不是让我们打表，而是每读入一个数重新计算一个数……吧：
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<cctype> 5 const int M = 5002; 6 7 int b[2][18000]; 8 9 inline void solve(const int &n) { 10 b[1][0] = 1; 11 int len = 1; 12 for(int i = 2; i <= n; ++i) { 13 int carry = 0, tmp; 14 for(int j = 0; j < len; ++j) { 15 tmp = carry + b[!(i & 1)][j] * i; 16 b[i & 1][j] = tmp % 10; 17 carry = tmp / 10; 18 } 19 while(carry) { 20 b[i & 1][len++] = carry % 10; 21 carry /= 10; 22 } 23 } 24 for(int j = len - 1; j >= 0; --j) 25 putchar(b[n & 1][j] + '0'); 26 puts(""); 27 } 28 29 template <typename T> 30 inline bool read(T &x) { 31 x = 0; 32 char ch = getchar(); 33 while(!isdigit(ch) && ch != EOF) ch = getchar(); 34 if(ch == EOF) return 0; 35 while(isdigit(ch)) { 36 x = x * 10 + (ch - '0'); 37 ch = getchar(); 38 } 39 return 1; 40 } 41 42 int main() { 43 int m; 44 while(read(m)) 45 solve(m); 46 return 0; 47 }
