The factorial of an integer N, written N!, is the product of all the integers from 1 through N inclusive. The factorial quickly becomes very large: 13! is too large to store in a 32-bit integer on most computers, and 70! is too large for most floating-point variables. Your task is to find the rightmost non-zero digit of n!. For example, 5! = 1 * 2 * 3 * 4 * 5 = 120, so the rightmost non-zero digit of 5! is 2. Likewise, 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040, so the rightmost non-zero digit of 7! is 4.
PROGRAM NAME: fact4
INPUT FORMAT
A single positive integer N no larger than 4,220.
SAMPLE INPUT (file fact4.in)
7
OUTPUT FORMAT
A single line containing but a single digit: the right most non-zero digit of N! .
SAMPLE OUTPUT (file fact4.out)
4
————————————————————————————
以为一个个把个位乘起来%10就好了，然并不，有些时候
例如75*4和74*14的最右非零位是不一样的，其实我们只需要手动去除2和5这两个质因子剩下的乘起来%10就可以了
纪念我的智障……
1 /* 2 ID: ivorysi 3 PROG: fact4 4 LANG: C++ 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 14 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 15 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 16 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 17 #define inf 0x7fffffff 18 #define MAXN 400005 19 #define ivorysi 20 #define mo 97797977 21 #define ha 974711 22 #define ba 47 23 #define fi first 24 #define se second 25 //#define pis pair<int,string> 26 using namespace std; 27 typedef long long ll; 28 int two,five; 29 int n; 30 int ans=1; 31 void divide(int &u) { 32 while(u%5==0) {++five;u/=5;} 33 while(u%2==0) {++two;u/=2;} 34 } 35 void solve() { 36 scanf("%d",&n); 37 siji(i,1,n) { 38 int tmp=i; 39 divide(tmp); 40 ans=(ans*tmp+10)%10; 41 } 42 two-=five; 43 siji(i,1,two) ans=ans*2%10; 44 printf("%d ",ans); 45 } 46 int main(int argc, char const *argv[]) 47 { 48 #ifdef ivorysi 49 freopen("fact4.in","r",stdin); 50 freopen("fact4.out","w",stdout); 51 #else 52 //freopen("f1.in","r",stdin); 53 #endif 54 solve(); 55 }