HDU 1042

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49365    Accepted Submission(s): 13859

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

123

 

Sample Output

126

 

Author

JGShining（极光炫影）

 

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思路：这里涉及到大数操作，下面分析一下大数阶乘的原理：

如果结果只有一位数，我们就存放在str[0]中。

如果结果有两位以上，就牵涉到进位操作。

str[0]存放结果的个位数，str[1]存放十位数，str[2]存放百位数,str[3]存放千位数。

具体存放状态如下：（结果倒数输出）

/*Title：HDU 1042Author：Dojking */#include <iostream>#include <string>using namespace std;int main(){int i, j, n, up, next;while (scanf("%d", &n) != EOF){string str;str += '1';     //将 str[0] 初始化为1 up = 0;         //up 表示进位 next = 0;       //next 表示乘以下一个数后的总结果 for (i = 2; i <= n; ++i){up = 0;for (j = 0; j < str.length(); ++j){next = (str[j]-'0') * i + up;   if (next > 9 && str[j+1] == '\0')  //最后一位需要进位，则在末尾追加一个值为0的空间 str += '0';str[j] = next % 10 + '0';          //当前位 up = next / 10;                    //进位 }}for (i = str.length()-1; i >= 0; --i)      //逆序输出 cout<<str[i];cout<<endl;}return 0;}