Beta Round ＃9 (酱油杯noi考后欢乐赛)乌鸦喝水

题目：http://www.contesthunter.org/contest/Beta%20Round%20%EF%BC%839%20%28%E9%85%B1%E6%B2%B9%E6%9D%AFnoi%E8%80%83%E5%90%8E%E6%AC%A2%E4%B9%90%E8%B5%9B%29/%E4%B9%8C%E9%B8%A6%E5%96%9D%E6%B0%B4

题解：真是一道神题！考场上绝对想不到标算orz！

题解写在代码注释里

代码：

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 100000+5

 #define maxm 500+100

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 #define for0(i,n) for(int i=0;i<=(n);i++)

 #define for1(i,n) for(int i=1;i<=(n);i++)

 #define for2(i,x,y) for(int i=(x);i<=(y);i++)

 #define for3(i,x,y) for(int i=(x);i>=(y);i--)

 #define mod 1000000007

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }
 int n,m,h,s[maxn],a[maxn],b[maxn],id[maxn];
 inline void add(int x,int y){for(;x<=n;x+=x&(-x))s[x]+=y;}
 inline int sum(int x){int t=;for(;x;x-=x&(-x))t+=s[x];return t;}
 inline bool cmp(int x,int y){return b[x]==b[y]?x<y:b[x]<b[y];}

 int main()

 {

     freopen("input","r",stdin);

     freopen("output.txt","w",stdout);

     n=read();m=read();h=read();
     for1(i,n)a[i]=read();
     for1(i,n)
     {
         int x=read();
         if(a[i]<=h)b[i]=(h-a[i])/x+;//b[i]表示该点最多可以被饮几次水
         if(b[i]>)add(i,);id[i]=i;
     }
     //可以证明，最后的答案一定是某个b[i]，所以我们按b[i]从小到大处理
     sort(id+,id+n+,cmp);
     int now=,pos=,ans=,st=n+;//now 表示当前的趟数
     //for1(i,n)cout<<i<<' '<<id[i]<<' '<<b[i]<<endl;
     for1(i,n)if(b[id[i]]){st=i;break;}
     for2(i,st,n)
     {
         while(now<=m)
         {
             int t=sum(n)-sum(pos);
             if(ans+t<b[id[i]])ans+=t,now++,pos=;//如果可以走完这一趟
             else break;
         }
         if(now>m)break;
         int l=pos,r=n,t=sum(pos);
         while(l<=r)
         {
             int mid=(l+r)>>;
             if(sum(mid)-t+ans>=b[id[i]])r=mid-;else l=mid+;
         }//二分到下一个点
         ans=b[id[i]];pos=l;
         for(;b[id[i]]==b[id[i+]];i++)add(id[i],-);//去除不能再次饮水的点
         add(id[i],-);
     }
     printf("%d\n",ans);

     return ;

 }