A PASCAL problem maximum square .

Let's build a house on vijos! Take a look at my solution:

The problem is the largest house that can be built, which is actually a classic problem of the dynamic rule, the largest sub-matrix.

In this problem, we can take the row as a stage, each unit of land is the state, and the optimal function f[i,j] refers to the maximum house unit that can be built with (i,j) as the vertex of the lower right corner of the square.

Seems a bit messy, for example, if the map is.

Then f is. f[1,1],f[1,2],f[2,1] are all 1 because they can only build a house with unit 1 with them at the vertex of the lower right corner of the square, while f[2,2] can build a house with unit 2 (the four vertices of the house are (1,1),(1,2),(2,1),(2,2)).

It is not difficult to infer that f[i,j] is closely related to f[i-1,j],f[i,j-1],f[i-1,j-1], and if f[i,j]=2, then f[i,j-1],f[i-1,j]f[i-1,j-1] must be greater than or equal to 1.

So the equation is.

f[i,j]=map[i,j]*(min+1);

At this point, the problem is solved.

0ms is no problem.

By the way, there are borders.

f[i,1]=map[i,1]

f[1,j]=map[1,j]

What's the question?? Illustrated? I don't understand.

Judge slowly. And then.. Output.. And then.. There is no then.

#include

#include

#include

#include

#include

using namespace std;

long long c=0,word[6]=,out=0;

if(s=="two"||s=="both"||s=="second")

if(s=="three"||s=="third")if(s=="four")

if(s=="five")

if(s=="six")

if(s=="seven")

if(s=="eight")

if(s=="nine")

if(s=="eleven")

if(s=="twelve")

if(s=="thirteen")

if(s=="fourteen")

if(s=="fifteen")

if(s=="sixteen")

if(s=="seventeen")

if(s=="eightteen")

if(s=="nineteen")

int main()

sort(word,word+c);

for(int i=0;icout

Can you take a look at an example?

This kind of myth is really ...

Rule (2) I don't understand.

It's the equivalent of a table.

0 1 2 3 4 5 6 7 （k）

x*x) can be extended indefinitely.

var n,m,i,j,k,t:longint;

a:array[0..1001,0..1001]of longint;

beginreadln(m,n);

n:=n*n mod 1000;

Ideas: 1. The first one is the starting point, which is the necessary point, so its value should be added to the accumulator;

2. In the future, you can only jump forward, and the rule of landing point is the largest of all the remaining points. For the study:

First time at 1 o'clock; The second time, choose the largest 3 points among the next 4 points and jump up; The third time, the greater of the remaining two points is selected and jumped up at 4 points; The fourth pick is the last 5 to jump up. Therefore: 1 5 4 3 13.

Embarrassment, I'm here to grab a job--I didn't debug forgive me.

vark:real;

a,b,n,i:longint;

beginreadln(n);

a:=1;b:=1;

for i:=1 to n do

begink:=k+a/b;

b:=a+b;

a:=b-a;

end;writeln(k:0:2);

end.

That should be right!

varch:char;

a:array[1..6] of char;

i,j:integer;

beginfor i:=1 to 5 do

a[i]:=chr(64+i);

for i:=1 to 5 do beginfor j:=1 to 5 do

write(a[j]);

writeln;

a[6]:=a[1];

for j:= 1 to 5 do

a[j]:=a[j+1];

end;end.

I tried! Pass! Used for!

This kind of question does not need to be looped.

The landlord does not know whether he is satisfied

program lt1；

beginwriteln('a b c d e');

writeln('b c d e a');

writeln('c d e a b');

writeln('d e a b c');

writeln('e a b c d');

end.They are all hand-beaten, and the watchtower lord adopts !!