The solution is a pre knowledge ： Difference array

Solution II pre knowledge ： Line segment tree

6 2 10

5 24

10 24

11 24

34 24

35 24

35 48

1
2

3
3

        The nature of the topic ： Each travel plan represents an interval . Each query is to query the current time x stay k After time （ namely x+k） How many intervals are included .

Solution 1 ： Difference array （ Best think 、 The least code solution ）

        The difference array is suitable for multiple interval modifications , Very few interval queries, or modify the interval first and then query the interval , This is not the case of querying while modifying . This problem uses a differential array , Each travel plan is the operation of the effective interval . Finally, prefix and Statistics , Then you can query .

/*
 Differential array version , Five minutes AC, It's much better than the line segment tree  
*/

#include <cstdio>
#define fi(i,l,r) for(int i=(l);i<=(r);++i)
#define fd(i,r,l) for(int i=(r);i>=(l);--i)

using namespace std;

int n,m,t;
int x,y;
int a[300005];

template<typename T>void Read(T &x)
{
	x=0;T k=1;char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') k=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	x*=k;
}
template<typename T>void Print(T x)
{
	if(x<0){putchar('-');Print(-x);return;}// This return Very important , Something can go wrong 
	if(x>9) Print(x/10);
	putchar((x%10)^48);
}

int main (void)
{
	Read(n);
	Read(m);
	Read(t);
	while(n--)
	{
		Read(x);
		Read(y);
		a[x-y<0?1:x-y+1]++;
		a[x+1]--;
	}
	fi(i,1,300004)
	  a[i]+=a[i-1];
	while(m--)
	{
		Read(x);
		Print(a[x+t]);
		putchar('\n');
	}
	return 0;
}

Solution 2 ： Line segment tree （ More yards ）

        Segment tree interval modification and interval query are in log Level , It also meets the requirements . This question only needs to write a single point query , Therefore, it is different from the general segment tree template .

#include <cstdio>
#define fi(i,l,r) for(int i=(l);i<=(r);++i)
#define fd(i,r,l) for(int i=(r);i>=(l);--i)
#define lc (rt<<1)
#define rc ((rt<<1)|1)
#define mid ((R+L)>>1)

using namespace std;
const int N=300005;

int n,m,t;
int x,y;
struct arr{
	int add,val;
}str[N<<2];// The line segment tree is usually four times larger  

template<typename T>void Read(T &x)
{
	x=0;T k=1;char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') k=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	x*=k;
}
template<typename T>void Print(T x)
{
	if(x<0){putchar('-');Print(-x);return;}// This return Very important , Something can go wrong 
	if(x>9) Print(x/10);
	putchar((x%10)^48);
}

void pushUp(int rt)
{
	str[rt].val=str[lc].val+str[rc].val;
}

void pushDown(int rt,int L,int R)
{
	str[lc].val+=str[rt].add*(mid-L+1);
	str[rc].val+=str[rt].add*(R-mid);
	str[lc].add+=str[rt].add;
	str[rc].add+=str[rt].add;
	str[rt].add=0;
}

void Upd(int rt,int L,int R,int ul,int ur,int addVal)
{
	if(L>ur||R<ul) return;
	if(ul<=L&&ur>=R)
	{
		str[rt].add+=addVal;
		str[rt].val+=addVal*(R-L+1);
		return;
	}
	if(str[rt].add!=0) pushDown(rt,L,R);
	if(ul<=mid) Upd(lc,L,mid,ul,ur,addVal);
	if(ur>mid) Upd(rc,mid+1,R,ul,ur,addVal);
	pushUp(rt);
}

int Query(int rt,int L,int R,int x)
{
	if(L==x&&R==x) return str[rt].val;
	if(str[rt].add!=0) {pushDown(rt,L,R);pushUp(rt);}
	if(x<=mid) return Query(lc,L,mid,x);
	else return Query(rc,mid+1,R,x);
}

int main (void)
{
	Read(n);
	Read(m);
	Read(t);
	while(n--)
	{
		Read(x);
		Read(y);
		Upd(1,1,300000,x-y<0?1:x-y+1,x,1);
	}
	while(m--)
	{
		Read(x);
		Print(Query(1,1,300000,x+t));
		putchar(10);
	}
	return 0;
}