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DescriptionKefa 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 friendto 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 leastd 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!InputThe 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 inorder to feel poor, respectively.Next n lines contain the descriptions of Kefa's friends, the(i + 1)-th line contains the description of thei-th friend of typemi,si (0 ≤ mi, si ≤ 109)— the amount of money and the friendship factor, respectively.OutputPrint the maximum total friendship factir that can be reached.Sample InputInput

4 5
75 5
0 100
150 20
75 1

Output

100

Input

5 1000 711 3299 1046 887 54

Output

111

两点，一点是爆数据，一点是求和线段

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<limits.h>#define zuida 100000using namespace std;int n,d,i;long long maxn,sum,fuzhu[zuida];struct myk{int m;int s;};myk a[zuida+10];bool cmp(myk a,myk b){    if(a.m<b.m)        return 1;    else        return 0;}int chazhao(int input){    int l,r,mid;    l=input;    r=n;    while(r-l>1)    {        mid=(l+r)/2;        if((a[mid].m-a[i].m)<d)            l=mid;        else            r=mid;    }    return l;}int main(void){    scanf("%d%d",&n,&d);    int j;    for(i=0;i<n;i++)        scanf("%d%d",&a[i].m,&a[i].s);    sort(a,a+n,cmp);    maxn=0;    fuzhu[0]=a[0].s;    for(i=1;i<n;i++)        fuzhu[i]=fuzhu[i-1]+a[i].s;    for(i=0;i<n;i++)    {        j=chazhao(i);        sum=fuzhu[j]-fuzhu[i]+a[i].s;        maxn=(sum>maxn)?sum:maxn;    }    printf("%I64d",maxn);    return 0;}