Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
The first line of the input contains two space-separated integers, n and d (1 ≤ n ≤ 105, 
) — the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 ≤ mi, si ≤ 109) — the amount of money and the friendship factor, respectively.
Print the maximum total friendship factir that can be reached.
4 5
75 5
0 100
150 20
75 1
100
5 100
0 7
11 32
99 10
46 8
87 54
111
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
题意：n个朋友 每个朋友拥有mon钱数 fri 友谊值
现在邀请若干朋友参加聚会 要求朋友间的钱数差小于d
输出被邀请参加聚会的 朋友的 友谊值的总和的最大值
题解：sort一下 记录前缀和 转换为取连续区间的和的最大值
固定左边界 按照 题目对钱数差距的要求 二分右边界 最后取区间和的max
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #define ll __int64 6 using namespace std; 7 ll n,d; 8 ll sum[100005]; 9 struct node 10 { 11 ll mon; 12 ll fri; 13 }N[100005]; 14 bool cmp(struct node aa,struct node bb) 15 { 16 if(aa.mon<bb.mon) 17 return true; 18 return false; 19 } 20 int main() 21 { 22 scanf("%I64d %I64d",&n,&d); 23 for(int i=1;i<=n;i++) 24 scanf("%I64d %I64d",&N[i].mon,&N[i].fri); 25 sort(N+1,N+1+n,cmp); 26 sum[0]=0; 27 ll ans=0; 28 for(int i=1;i<=n;i++) 29 sum[i]=sum[i-1]+N[i].fri; 30 for(int i=1;i<=n;i++) 31 { 32 ll l=i,r=n,mid; 33 while(l<r) 34 { 35 36 mid=(l+r+1)>>1; 37 38 if((N[mid].mon-N[i].mon)>=d) 39 r=mid-1; 40 else 41 l=mid; 42 } 43 //cout<<i<<" "<<l<<" "<<mid<<" "<<r<<"&&&"<<endl; 44 ans=max(ans,sum[l]-sum[i-1]); 45 } 46 cout<<ans<<endl; 47 return 0; 48 }