题解：忽然想起来我好像还没写过高精度除以单精度，于是拿这题练练手。。。没想到1A了。。。
代码：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<string> 8 #include<set> 9 #include<map> 10 #include<vector> 11 #include<queue> 12 #define maxn (5000+10) 13 #define for0(i,n) for(int i=0;i<=n;i++) 14 #define for1(i,n) for(int i=1;i<=n;i++) 15 #define for2(i,s,t) for(int i=s;i<=t;i++) 16 #define for3(i,t,s) for(int i=t;i>=s;i--) 17 #define ll long long 18 #define mod 10000 19 using namespace std; 20 inline int read() 21 { 22 int x=0,f=1;char ch=getchar(); 23 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 24 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 25 return x*f; 26 } 27 class bigg 28 { 29 public: 30 int c[maxn],l; 31 bigg(){memset(c,0,sizeof(c));l=0;} 32 bigg operator = (int b) 33 { 34 memset(c,0,sizeof(c));l=0; 35 while(b)c[++l]=b%mod,b/=mod; 36 return(*this); 37 } 38 bigg operator =(const bigg &b) 39 { 40 l=b.l; 41 for1(i,l)c[i]=b.c[i]; 42 return(*this); 43 } 44 bigg operator +(const bigg &b) 45 { 46 bigg a; 47 a.l=max(l,b.l); 48 for1(i,l) 49 { 50 a.c[i]+=a.l+b.l; 51 a.c[i+1]=a.c[i]/mod; 52 a.c[i]%=mod; 53 } 54 if(c[a.l+1])a.l++; 55 return a; 56 } 57 bigg operator *(int b) 58 { 59 bigg a;a.l=l; 60 int x=0; 61 for1(i,l) 62 { 63 x+=b*c[i]; 64 a.c[i]=x%mod; 65 x/=mod; 66 } 67 while(x)a.c[++a.l]=x%mod,x/=mod; 68 return a; 69 } 70 bigg operator /(int b) 71 { 72 bigg a; 73 int x=0; 74 for3(i,l,1) 75 { 76 x=x*mod+c[i]; 77 a.c[i]=x/b; 78 x%=b; 79 } 80 a.l=l; 81 while(!a.c[a.l])a.l--; 82 return a; 83 } 84 void print() 85 { 86 printf("%d",c[l]); 87 for3(i,l-1,1)printf("%04d",c[i]);printf(" "); 88 } 89 }; 90 int n,a[maxn],b[maxn],c[maxn]; 91 inline bool cmp(int x,int y){return a[x]*b[x]<a[y]*b[y];} 92 int main() 93 { 94 freopen("input.txt","r",stdin); 95 freopen("output.txt","w",stdout); 96 n=read()+1; 97 for1(i,n)a[i]=read(),b[i]=read(),c[i]=i; 98 sort(c+1,c+n+1,cmp); 99 bigg ans;ans=1; 100 for1(i,n-1)ans=ans*a[c[i]]; 101 ans=ans/b[c[n]]; 102 ans.print(); 103 return 0; 104 }