N!
Time Limit: 10000/5000 MS (java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 79143 Accepted Submission(s): 23137
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.
Sample Input 1 2 3
Sample Output 1 2 6
大數(shù)的階乘，10000的階乘為35000多位數(shù)。 開個3000的數(shù)組，數(shù)組里的每個數(shù)儲存13位數(shù)，就足夠了 這算是10萬億進制了吧233333 代碼：
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 9999using namespace std;long long a[10005][3001];void deal(){ MST(a, 0); a[0][1] = 1; for (int i = 1; i <= 10000; i++) { long long r = 0; for (int j = 1; j <= 3000; j++) { a[i][j] = a[i - 1][j] * i + r; if (a[i][j] < 0) printf("/n"); r = 0; if (a[i][j] > 10000000000000) { r = a[i][j] / 10000000000000; a[i][j] %= 10000000000000; } } }}int main(){ deal(); int n; while (cin >> n) { int i = 3000; while (a[n][i] == 0)i--; printf("%lld", a[n][i] ); // 剛開始不用補0 i--; for (; i >= 1; i--) printf("%013lld", a[n][i] ); // 不夠13位就補0 printf("/n"); }}新聞熱點
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