loj 10005

题目分析：

• 每次选最小的两个合并，最后得到最大的数
• 每次选最大的两个合并，最后得到最小的数
• 用优先队列实现

Code:

#include <bits/stdc++.h>
using namespace std;
#define maxn 50010

int n,a[maxn];
long long maxd,mind;
struct hep_min {
	long long x;
	bool operator < (const hep_min& rhs) const {
		return x > rhs.x;
	}
};

inline int read_() {
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}

inline void work_min_() {
	priority_queue<long long >q;
	for(int i=1;i<=n;++i) q.push( (long long) a[i]);
	long long x,y;
	while(!q.empty()) {
		if(q.size()==1) {
			mind=q.top();
			return;
		}
		x=q.top();q.pop();
		y=q.top();q.pop();
		q.push(x*y+1);
	}
}

inline void work_max_() {
	priority_queue<hep_min> q;
	for(int i=1;i<=n;++i) {
		q.push( (hep_min) { (long long ) a[i] } );
	} 
	long long x,y;
	while(!q.empty()) {
		if(q.size()==1) {
			maxd=q.top().x;
			return;
		}
		x=q.top().x;q.pop();
		y=q.top().x;q.pop();
		q.push( (hep_min) {x*y+1} );
	}
}

void readda_() {
	n=read_();
	for(int i=1;i<=n;++i) a[i]=read_();
	int x=read_();
	work_max_();
	work_min_();
	printf("%lld",maxd-mind);
}

int main() {
	freopen("a.txt","r",stdin);
	readda_();
	return 0;
}