题面描述
求 (1! imes 2! imes 3! imes ... imes n!) 末尾 (0) 的个数
题解
其实原题的数据范围不大，(n≤1e8) 但给了我很多思考
设 (F(i)= sum_{i=1}^{infty}leftlfloordfrac{n}{5^i} ight floor)
则
[ans=sum_{i=1}^{n} F(i)= sum_{i=1}^{n}sum_{j=1}^{infty}leftlfloordfrac{n}{5^i}
ight
floor=sum_{j=1}^{infty}sum_{i=1}^{n}leftlfloordfrac{n}{5^i}
ight
floor=sum_{j=1}^{infty}(5^j imes sum_{i=1}^{lfloorfrac{n}{5^i}
floor-1}i+ lfloorfrac{n}{5^i}
floor imes(n%5^j+1)
]
柿子有点长
代码实现
#include<bits/stdc++.h>
using namespace std;
long long N,ans;
struct IO{
static const int S=1<<21;
char buf[S],*p1,*p2;int st[105],Top;
~IO(){clear();}
inline void clear(){fwrite(buf,1,Top,stdout);Top=0;}
inline void pc(const char c){Top==S&&(clear(),0);buf[Top++]=c;}
inline char gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
inline IO&operator >> (char&x){while(x=gc(),x==' '||x=='
'||x=='r');return *this;}
template<typename T>inline IO&operator >> (T&x){
x=0;bool f=0;char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-') f^=1;ch=gc();}
while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=gc();
f?x=-x:0;return *this;
}
inline IO&operator << (const char c){pc(c);return *this;}
template<typename T>inline IO&operator << (T x){
if(x<0) pc('-'),x=-x;
do{st[++st[0]]=x%10,x/=10;}while(x);
while(st[0]) pc('0'+st[st[0]--]);return *this;
}
}fin,fout;
int main(){
fin>>N;
for(int j=5;j<=N;j*=5){
ans+=j*(N/j)*(N/j-1)>>1;
ans+=(N/j)*(N%j+1);
}
fout<<ans;
return 0;
}