思路：

iwtwiioi大爷题解：http://www.cnblogs.com/iwtwiioi/p/4007263.html

注意是四舍五入

//By SiriusRen#include 
     
      using namespace std;typedef long long ll;const int N=1000500;ll n,d,a[N],X[N],sum[N],top;struct Point{ll x,y;}point[N],q[N],ask;double Ans,ans;Point operator-(Point a,Point b){    Point c;c.x=a.x-b.x,c.y=a.y-b.y;return c;}double cross(Point a,Point b,Point c){    a=a-c,b=b-c;return a.x*b.y-a.y*b.x;}int main(){    scanf("%lld%lld",&n,&d);    for(int i=1;i<=n;i++){        scanf("%lld%lld",&a[i],&X[i]);        sum[i]=sum[i-1]+a[i];        point[i].x=d*i,point[i].y=sum[i-1];    }    for(int i=1;i<=n;i++){        while(top>1&&cross(point[i],q[top],q[top-1])>=0)top--;        q[++top]=point[i];        ask.x=X[i]+d*i,ask.y=sum[i];        int l=1,r=top,lmid,rmid;        while(r-l>=3){            lmid=l+(r-l)/3,rmid=r-(r-l)/3;            double k1=(double)(q[lmid].y-ask.y)/(q[lmid].x-ask.x);            double k2=(double)(q[rmid].y-ask.y)/(q[rmid].x-ask.x);            if(k2>k1)l=lmid;            else r=rmid;        }ans=0;        for(int i=l;i<=r;i++)ans=max(ans,(double)(q[i].y-ask.y)/(q[i].x-ask.x));        Ans+=ans;    }    printf("%.0lf\n",Ans);}