UVa OJ 623 阶乘

最新推荐文章于 2021-04-08 20:46:26 发布

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#AOAPC_1 #阶乘

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本文介绍了一种使用数组存储和计算大数阶乘的方法，解决了传统整型数据无法表示超长数字的问题。通过逐位相乘和进位处理，实现了高效的大数阶乘计算。

UVa OJ 623

500！

In these days you can more and more often happen to see programs which perform some useful calculations being executed rather then trivial screen savers. Some of them check the system message queue and in case of finding it empty (for examples somebody is editing a file and stays idle for some time) execute its own algorithm.

As an examples we can give programs which calculate primary numbers.

One can also imagine a program which calculates a factorial of given numbers. In this case it is the time complexity of order O(n) which makes troubles, but the memory requirements. Considering the fact that 500! gives 1135-digit number no standard, neither integer nor floating, data type is applicable here.

Your task is to write a programs which calculates a factorial of a given number.

Assumptions: Value of a number ``n" which factorial should be calculated of does not exceed 1000 (although 500! is the name of the problem, 500! is a small limit).

Input

Any number of lines, each containing value ``n" for which you should provide value of n!

Output

2 lines for each input case. First should contain value ``n" followed by character `!'. The second should contain calculated value n!.

Sample Input

Sample Output

10!
3628800
30!
265252859812191058636308480000000
50!
30414093201713378043612608166064768844377641568960512000000000000
100!
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

用数组来保存运算结果。为了方便起见，我们用f[0]保存个位，f[1]保存十位...（为什么要逆序表示？方便进位），每次模拟手算来计算n!。在输出时要忽略前导0。注意，如果结果本身就是0，那么忽略前导零将会导致什么都不输出。幸好n!肯定不等于0.

#include <cstdio>
#include <cstdlib>
#include <cstring> 

using namespace std;

const int MAXN = 3000;
int f[MAXN];

bool foo(int n, int f[])
{
    if (n < 0)
    {
        printf("n must be positive!\n");
        return false;
    }
    else
    {
        f[0] = 1;
        for (int i = 2; i <= n; ++i)
        {
            int c = 0;    // c是进位 
            for (int j = 0; j < MAXN; ++j)
            {
                int s = f[j] * i + c;
                f[j] = s % 10;
                c = s / 10;
            }
        }
        int end = MAXN-1;
        for (; end >= 0; --end)
        {
            if (f[end]) break;
        }
        printf("%d!\n", n);
        for (int i = end; i >= 0; --i)
        {
            printf("%d", f[i]);
        }
        printf("\n");
    }
    return true;
}

int main(int argc, char *argv[])
{
    int n = 0;
    while (scanf("%d", &n) != EOF)
    {
        memset(f, 0, sizeof(f));
        foo(n, f);
    }
    return 0;
}