Codeforces Round #181 (Div. 2) E. Empire Strikes Back N!∣∏K,i=1ai!

http://codeforces.com/contest/300/problem/E

题目描述：

求最小正整数 N ，使得 N!∣∏Ki=1ai! 。

记 sum=∏Ki=1ai!。

首先可以想到，对于每一个质数 pi ，找到最小的 Ni 使得 Ni! 中质因子 pi 的出现次数大于等于 sum 里的。

这个玩意儿显然可以二分，于是问题转变成如何求一个数的阶乘里有多少个 pi 因子，答案是 ∑⌊Nipki⌋ ，这部分的复杂度是 O(log2K) 。

再考虑如何求 sum 里质因子 pi 的个数。

令 numx 表示 sum 中因子 x 的出现次数。

如果不考虑阶乘里面的合数分解出来的因子的话，可以简单的求一下后缀和（详见代码），而之后对于一个合数 k ，我们可以用 numk 去更新 numprk 和 numkprk 的值，其中 prk 表示 k 的最质数因子。这样的话每次总是用大数去更新小数，因此从大到小递推即可。

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,1,-1,-1,1};
//const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e6+10;
const int maxx=1e7+10+6e6;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=10086;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}

LL d[maxx],pr[maxx],tot;
bool vis[maxx];
void init()
{
for(LL i = 2; i < maxx; ++i)
{
if(!d[i])
pr[tot++] = d[i] = i;
for(LL j = 0, k; (k = i * pr[j]) < maxx; ++j)
{
d[k] = pr[j];
if(d[i] == pr[j])
break;
}
}
me(vis,1);
for(int i=0;i<tot;i++)
if(vis[pr[i]]) vis[pr[i]]=0;
}
int n;
LL num[maxx];
void solve()
{
s_1(n);
LL sum=0;
LL mx=-1;
FOR(1,n,i)
{
LL x;
S_1(x);
sum+=x;
mx=max(mx,x);
num[x]++;
}
fOR(mx,2,i) num[i-1]+=num[i];
fOR(mx,2,i) if(vis[i])
num[d[i]]+=num[i],num[i/d[i]]+=num[i];
LL ans=1;
for(int i=0;i<tot&&num[pr[i]];i++)
{
LL l=1,r=sum,k;
W(l<=r)
{
LL mid=(l+r)>>1;
LL cnt=0;
for(LL tmp=mid;tmp;tmp/=pr[i])
cnt+=tmp/pr[i];
if(cnt<num[pr[i]]) l=mid+1;
else r=(k=mid)-1;
}
ans=max(ans,k);
}
print(ans);
}
int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d:\n",cas);
solve();
}
}