單個詢問二分答案再二維BIT即可，多組詢問直接整體二分。註意保證復雜度。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 510
#define M 60010
char getc(){char c=getchar();while ((c<‘A‘||c>‘
 
Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3
 
)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N][N],ans[M],tree[N][N],t;
struct data{int xl,yl,xr,yr,k,i;
}q[M],p[M];
struct data2
{
    int x,y,k;
    bool operator <(const data2&a) const
    {
        return k<a.k;
    }
}b[N*N];
int query(int x,int y)
{
    int s=0;
    for (int i=x;i;i^=i&-i)
        
 
for (int j=y;j;j^=j&-j)
        s+=tree[i][j];
    return s;
}
void add(int x,int y,int p)
{
    for (int i=x;i<=n;i+=i&-i)
        for (int j=y;j<=n;j+=j&-j)
        tree[i][j]+=p;
}
int query(int xl,int yl,int xr,int yr){return query(xr,yr)-query(xl-1,yr)-query(xr,yl-1)+query(xl-1,yl-1);}
void solve(int l,int r,int x,int y)
{
    if (l>r) return;
    if (x==y)
    {
        for (int i=l;i<=r;i++) ans[q[i].i]=b[x].k;
        return;
    }
    int mid=x+y>>1,L=l,R=r;
    for (int i=x;i<=mid;i++) add(b[i].x,b[i].y,1);
    for (int i=l;i<=r;i++)
    {
        int u=query(q[i].xl,q[i].yl,q[i].xr,q[i].yr);
        if (u<q[i].k) q[i].k-=u,p[R--]=q[i];
        else p[L++]=q[i];
    }
    for (int i=l;i<=r;i++) q[i]=p[i];
    for (int i=x;i<=mid;i++) add(b[i].x,b[i].y,-1);
    solve(l,L-1,x,mid);
    solve(R+1,r,mid+1,y);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("matrix.in","r",stdin);
    freopen("matrix.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            a[i][j]=read();
            t++;b[t].k=a[i][j],b[t].x=i,b[t].y=j;
        }
    sort(b+1,b+t+1);
    for (int i=1;i<=m;i++) q[i].xl=read(),q[i].yl=read(),q[i].xr=read(),q[i].yr=read(),q[i].k=read(),q[i].i=i;
    solve(1,m,1,n*n);
    for (int i=1;i<=m;i++) printf("%d\n",ans[i]);
    return 0;
}

BZOJ2738 矩陣乘法（整體二分+樹狀數組）