| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
就是一个普通的分解质因数吧...自己真的是没救了...就是把从a1到a10全部分解质因数,然后所有质因数+1相乘,不停的%10
我做法是跟筛素数一样,从2开始,判断每个素数因子有多少,然后+1*现在的答案,因为只考虑最后一位,那么不停的%10
#include <stdio.h>
int i,k,n,j,s[12];
int main()
{
n=1;
for(i=0;i<10;i++) scanf("%d",&s[i]);
for(i=2;i<100;i++)
{
k=0;
for(j=0;j<10;j++)
{
while(s[j]%i==0)
{
k++;
s[j]/=i;
}
}
n=(n*k+n)%10;
}
k=0;
for(j=0;j<10;j++)
if(s[j]>1) k++;
n=(n*k+n)%10;
printf("%d\n",n);
return(0);
}
Followed by:
Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator