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：
1 #include<cstdio> 2 3 #include<cstdlib> 4 5 #include<cmath> 6 7 #include<cstring> 8 9 #include<algorithm> 10 11 #include<iostream> 12 13 #include<vector> 14 15 #include<map> 16 17 #include<set> 18 19 #include<queue> 20 21 #include<string> 22 23 #define inf 1000000000 24 25 #define maxn 100000+5 26 27 #define maxm 500+100 28 29 #define eps 1e-10 30 31 #define ll long long 32 33 #define pa pair<int,int> 34 35 #define for0(i,n) for(int i=0;i<=(n);i++) 36 37 #define for1(i,n) for(int i=1;i<=(n);i++) 38 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42 43 #define mod 1000000007 44 45 using namespace std; 46 47 inline int read() 48 49 { 50 51 int x=0,f=1;char ch=getchar(); 52 53 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 54 55 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 56 57 return x*f; 58 59 } 60 int n,m,h,s[maxn],a[maxn],b[maxn],id[maxn]; 61 inline void add(int x,int y){for(;x<=n;x+=x&(-x))s[x]+=y;} 62 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=s[x];return t;} 63 inline bool cmp(int x,int y){return b[x]==b[y]?x<y:b[x]<b[y];} 64 65 int main() 66 67 { 68 69 freopen("input","r",stdin); 70 71 freopen("output.txt","w",stdout); 72 73 n=read();m=read();h=read(); 74 for1(i,n)a[i]=read(); 75 for1(i,n) 76 { 77 int x=read(); 78 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 79 if(b[i]>0)add(i,1);id[i]=i; 80 } 81 //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 82 sort(id+1,id+n+1,cmp); 83 int now=1,pos=0,ans=0,st=n+1;//now 表示当前的趟数 84 //for1(i,n)cout<<i<<' '<<id[i]<<' '<<b[i]<<endl; 85 for1(i,n)if(b[id[i]]){st=i;break;} 86 for2(i,st,n) 87 { 88 while(now<=m) 89 { 90 int t=sum(n)-sum(pos); 91 if(ans+t<b[id[i]])ans+=t,now++,pos=0;//如果可以走完这一趟 92 else break; 93 } 94 if(now>m)break; 95 int l=pos,r=n,t=sum(pos); 96 while(l<=r) 97 { 98 int mid=(l+r)>>1; 99 if(sum(mid)-t+ans>=b[id[i]])r=mid-1;else l=mid+1; 100 }//二分到下一个点 101 ans=b[id[i]];pos=l; 102 for(;b[id[i]]==b[id[i+1]];i++)add(id[i],-1);//去除不能再次饮水的点 103 add(id[i],-1); 104 } 105 printf("%d ",ans); 106 107 return 0; 108 109 }