একটা ম্যাট্রিক্স দেওয়া থাকবে N*N গ্রিড এর
বের করা লাগবে
(I, J) to (I+S-1, J+S-1)
এইটার মধ্যে ম্যাক্সিমাম মান কে?
উদাহরণঃ
ইনপুটঃ
#include <bits/stdc++.h>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cout<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
int n,q;
int ara[505][505];
int sparse[505][505][16];
void build()
{
for(int k=0; (1<<k)<=n; k++)
{
for(int i=0; i+(1<<k)-1<n; i++)
for(int j=0; j+(1<<k)-1<n; j++)
{
if(k==0)
sparse[i][j][k]=ara[i][j];
else
{
int a=1<<(k-1);
sparse[i][j][k]=max(max(sparse[i][j][k-1],sparse[i+a][j][k-1]),max(sparse[i][j+a][k-1],sparse[i+a][j+a][k-1]));
}
}
}
}
int main()
{
int t;
sf(t);
TEST_CASE(t)
{
sff(n,q);
loop(i,n) loop(j,n) sf(ara[i][j]);
build();
PRINT_CASE;
while(q--)
{
int i,j,s;
sfff(i,j,s);
int k=log2(s);
int a=1<<k;
i--;
j--;
int ans=max(max(sparse[i][j][k],sparse[i+s-a][j][k]),max(sparse[i][j+s-a][k],sparse[i+s-a][j+s-a][k]));
pf("%d\n",ans);
}
}
return 0;
}
বের করা লাগবে
(I, J) to (I+S-1, J+S-1)
এইটার মধ্যে ম্যাক্সিমাম মান কে?
উদাহরণঃ
ইনপুটঃ
1
4 5
67 1 2 3
8 88 21 1
89 12 0 12
5 5 5 5
1 1 2
1 3 2
3 3 2
1 1 4
2 2 3
OUTPUT:
Case 1:
88
21
12
89
88
Case 1:
88
21
12
89
88
#include <bits/stdc++.h>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cout<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
int n,q;
int ara[505][505];
int sparse[505][505][16];
void build()
{
for(int k=0; (1<<k)<=n; k++)
{
for(int i=0; i+(1<<k)-1<n; i++)
for(int j=0; j+(1<<k)-1<n; j++)
{
if(k==0)
sparse[i][j][k]=ara[i][j];
else
{
int a=1<<(k-1);
sparse[i][j][k]=max(max(sparse[i][j][k-1],sparse[i+a][j][k-1]),max(sparse[i][j+a][k-1],sparse[i+a][j+a][k-1]));
}
}
}
}
int main()
{
int t;
sf(t);
TEST_CASE(t)
{
sff(n,q);
loop(i,n) loop(j,n) sf(ara[i][j]);
build();
PRINT_CASE;
while(q--)
{
int i,j,s;
sfff(i,j,s);
int k=log2(s);
int a=1<<k;
i--;
j--;
int ans=max(max(sparse[i][j][k],sparse[i+s-a][j][k]),max(sparse[i][j+s-a][k],sparse[i+s-a][j+s-a][k]));
pf("%d\n",ans);
}
}
return 0;
}
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