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  • 【ACM】hdu_1042_N!_201308071639

    N!
    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 43937    Accepted Submission(s): 12362


    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
     


    #include <stdio.h>
    #include <string.h>
    #define MAX_LEN 40000
    int s[MAX_LEN];

    int main()
    {
        int n;
        while(scanf("%d",&n)!=EOF)
        {
            int i,j,t,sum;
            memset(s,0,sizeof(s));
            s[0]=1;
            for(i=2;i<=n;i++)
            {
                t=0;;
                for(j=0;j<MAX_LEN;j++)
                {
                    sum=s[j]*i+t;
                    s[j]=sum%10;
                    t=sum/10;
                }
            }
            for(i=MAX_LEN-1;(i>0)&&(s[i]==0);i--);
            for(;i>=0;i--)
            printf("%d",s[i]);
            printf(" ");
        }
        return 0;   
    }

    //此解法耗时3500MS


    #include <stdio.h>
    #include <string.h>
    #define MAX_LEN 40000
    int a[MAX_LEN];
    int main()
    {
        int m;
        while(scanf("%d",&m)!=EOF)
        {
            int i,j,len=0,tem,jin;
            memset(a,0,sizeof(a));
            a[0]=1;
            for(i=2;i<=m;i++)
            {
                jin=0;
                for(j=0;j<=len;j++)
                {
                    tem=a[j]*i+jin;
                    a[j]=tem%10;
                    jin=tem/10;
                    if(j==len&&jin!=0)  //如果需要进一位,len+1,这样可以减少循环次数
                    len++;
                    }
                }
                for(i=len;i>=0;i--)
                printf("%d",a[i]);
                printf(" ");
            }
            return 0;
    }

    //此解法耗时1218MS

    //此解法节省时间

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  • 原文地址:https://www.cnblogs.com/xl1027515989/p/3243650.html
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