题意:给出一个长度为 n的数列 a和一个长度为 m 的数列 b，求 a有多少个长度为 m的连续子数列能与 b匹配。两个数列可以匹配，当且仅当存在一种方案，使两个数列中的数可以两两配对，两个数可以配对当且仅当它们的和不小于 h。

题解:先把b排序,要想能匹配,由hall定理,b的每个子集(大小为x)都至少有x条连向b,bi递增,和bi连的边也递增,那么当bi连边大于等于i时即可,所以当min(bi-i)>=0时满足条件

线性扫一遍即可,每个a二分b更新线段树即可

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000007

#define ld long double

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

//#define cd complex<double>

#define ull unsigned long long

//#define base 1000000000000000000

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

template<typename T>

inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>

inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=150000+10,maxn=50000+10,inf=0x3f3f3f3f;

int mi[N<<2],lazy[N<<2],a[N],b[N];

void pushup(int rt)

{

    mi[rt]=MIN(mi[rt<<1],mi[rt<<1|1]);

}

void pushdown(int rt)

{

    if(lazy[rt]!=0)

    {

        mi[rt<<1]+=lazy[rt];

        mi[rt<<1|1]+=lazy[rt];

        lazy[rt<<1]+=lazy[rt];

        lazy[rt<<1|1]+=lazy[rt];

        lazy[rt]=0;

    }

}

void build(int l,int r,int rt)

{

    lazy[rt]=0;

    if(l==r){mi[rt]=-l;return ;}

    int m=(l+r)>>1;

    build(ls),build(rs);

    pushup(rt);

}

void update(int L,int R,int x,int l,int r,int rt)

{

    if(L<=l&&r<=R)

    {

        mi[rt]+=x;

        lazy[rt]+=x;

        return ;

    }

    int m=(l+r)>>1;

    pushdown(rt);

    if(L<=m)update(L,R,x,ls);

    if(m<R)update(L,R,x,rs);

    pushup(rt);

}

int main()

{

    int n,m,h;

    scanf("%d%d%d",&n,&m,&h);

    for(int i=1;i<=m;i++)scanf("%d",&b[i]);

    for(int i=1;i<=n;i++)scanf("%d",&a[i]);

    sort(b+1,b+1+m);

    build(1,m,1);

    int ans=0;

    for(int i=1;i<=m;i++)

    {

        int p=lower_bound(b+1,b+1+m,h-a[i])-b;

//        printf("%d %d ----\n",i,p);

        if(p<=m)update(p,m,1,1,m,1);

        if(mi[1]>=0)ans++;

    }

    for(int i=m+1;i<=n;i++)

    {

        int p=lower_bound(b+1,b+1+m,h-a[i])-b;

        if(p<=m)update(p,m,1,1,m,1);

        p=lower_bound(b+1,b+1+m,h-a[i-m])-b;

        if(p<=m)update(p,m,-1,1,m,1);

        if(mi[1]>=0)ans++;

    }

    printf("%d\n",ans);

    return 0;

}

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