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=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';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=0;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+1;//b[i]表示该点最多可以被饮几次水
        if(b[i]>0)add(i,1);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+1,id+n+1,cmp);
    int now=1,pos=0,ans=0,st=n+1;//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=0;//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)>>1;
            if(sum(mid)-t+ans>=b[id[i]])r=mid-1;else l=mid+1;
        }//二分到下一个点
        ans=b[id[i]];pos=l;
        for(;b[id[i]]==b[id[i+1]];i++)add(id[i],-1);//去除不能再次饮水的点
        add(id[i],-1);
    }
    printf("%d
",ans);

    return 0;

}