国王游戏sol

国王游戏
贪心策略较为简单，可以邻项考虑，式子化简后按照x*y排序即可
要用到高精度乘除。。

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define SZ(x) ((int)x.size())
#define ALL(x) x.begin(),x.end()
#define U(i,u) for(register ll i=head[u];i;i=nxt[i])
#define rep(i,a,b) for(register ll i=(a);i<=(b);++i)
#define per(i,a,b) for(register ll i=(a);i>=(b);--i)
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> PII;
typedef vector<int> VI;
template<class T> inline void read(T &x){
	x=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c)){x=x*10+c-'0';c=getchar();}x*=f;
}
template<class T> inline void cmin(T &x, T y){x=x<y?x:y;}
template<class T> inline void cmax(T &x, T y){x=x>y?x:y;}
const int N=200001;
int n;
int tim[N],ans[N],mx[N],tmp[N];
struct node{
	int x,y;
}d[N];
bool cmp(node a,node b){
	return a.x*a.y<b.x*b.y;
}
void mul(int a[],int b){
	int len1=a[0],len2=len1;
	memset(tmp,0,sizeof(tmp));
	rep(i,1,len1){
		tmp[i]+=a[i]*b;
		tmp[i+1]+=tmp[i]/10;
		tmp[i]%=10;
	}
	while(tmp[len2+1]>0){
		++len2;
		tmp[len2+1]+=tmp[len2]/10;
		tmp[len2]%=10;
	}
	tim[0]=len2;rep(i,1,len2){tim[i]=tmp[i];}
}
void div(int a[],int b,int c[]){
	memset(ans,0,sizeof(ans));
	int len=a[0],d=0;per(i,len,1){
		c[i]=(d*10+a[i])/b;
		d=(d*10+a[i])%b;
		if(c[0]==0&&c[i]!=0)c[0]=i;
	}
	ans[0]=c[0];rep(i,1,c[0])ans[i]=c[i];
}
bool compare(int a[],int b[]){
	if(a[0]==b[0]){
		per(i,a[0],1){
			if(a[i]>b[i])return 1;
			if(a[i]<b[i])return 0;
		}
	}
	if(a[0]>b[0])return 1;
	else if(a[0]<b[0])return 0;
	else return 0;
}
void cp(){
	memset(mx,0,sizeof(mx));
	mx[0]=ans[0];rep(i,1,mx[0]){
		mx[i]=ans[i];
	}
}
int main(){
	read(n);rep(i,1,n+1){
		read(d[i].x);read(d[i].y);
	}
	sort(d+2,d+n+2,cmp);
	tim[0]=1;tim[1]=1;
	mul(tim,d[1].x);
	rep(i,2,n+1){		
		div(tim,d[i].y,ans);
		if(compare(ans,mx)){
			cp();
		}
		mul(tim,d[i].x);
	}
	per(i,mx[0],1){
		printf("%d",mx[i]);
	}
	return 0;
}