HDU 1042 N！ N的阶乘（大数）
发布时间：2021-01-28 16:34:28 所属栏目：大数据 来源：网络整理
导读：N! Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 75303????Accepted Submission(s): 21981 Problem Description Given an integer N(0 ≤ N ≤ 10000),your task is to calculate N! ? In
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N!Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 75303????Accepted Submission(s): 21981 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? Author 求n的阶乘嘛！ 基础题，依然WA，也是够了！ #include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define N 1100000
int a[N];
int main()
{
int i,j,n,count,t,k;
while(scanf("%d",&n)!=EOF)
{
a[0]=1;
count=1;
for(i=1;i<=n;i++)
{
k=0;
for(j=0;j<count;j++)
{
t=a[j]*i+k;
a[j]=t%10;
k=t/10;
}
while(k)
{
a[count++]=k%10;
k=k/10;
}
}
for(i=count-1;i>=0;i--)
{
printf("%d",a[i]);
}
printf("n");
}
return 0;
}
10000的阶乘也就35660位，要是担心TLE，或者超限的话，也可以按10000进制来写： #include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define N 1100000
int a[N];
int main()
{
int i,&n)!=EOF)
{
a[0]=1;
count=1;
for(i=1;i<=n;i++)
{
k=0;
for(j=0;j<count;j++)
{
t=a[j]*i+k;
a[j]=t%10000;
k=t/10000;
}
while(k)
{
a[count++]=k%10000; ///这里以10000为进位单位
k=k/10000;
}
}
printf("%d",a[count-1]); ///注意一下，第一个数不是“%04d”；
for(i=count-2;i>=0;i--)
{
printf("%04d",a[i]); ///其他的按四位的输出，不足补零
}
printf("n");
}
return 0;
}
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