Hello everyone, I am Bubble. Today I bring you the content of the review. Because of the review, I will pass it by.
content
1. Square sequence
Topic link: Square Sequence - Lanqiao Cloud Course (lanqiao.cn)
Topic requirements:
Xiao Ming wants to find two positive integers XX and YY, satisfying
- 2019<X<Y;
- 2019^2, X^2, Y^2 form an arithmetic sequence.
What is the minimum value of X + YX + Y among all possible solutions?
Problem solving ideas:
Violence
#include<bits/stdc++.h>
using namespace std;
int main()
{
for(int i=2020;i<10001;i++)
{
int x = i*i*2-2019*2019;
int y = sqrt(x);
if(y*y==x)
{
cout<<i+y;
return 0;
}
}
}2. Crossing the minefield
Topic link: Crossing the minefield - Lanqiao Cloud Course (lanqiao.cn)
Topic requirements:
X-star's tank tank is very strange, it must alternately pass through the positive energy radiation area and the negative energy radiation area to maintain normal operation, otherwise it will be scrapped.
A tank needs to go from area A to area B (areas A and B are safe areas, with no positive or negative energy characteristics), how to get the shortest path?
The known map is a square matrix, on which the A and B areas are marked with letters, and the other areas are marked with positive or negative signs to indicate the positive and negative energy radiation areas, respectively.
E.g:
A + - + -
- + - - +
- + + + -
+ - + - +
B + - + -
Tanks can only move horizontally or vertically to adjacent zones.
Problem solving ideas:
BFS template
#include<bits/stdc++.h>
using namespace std;
const int N = 101;
char s[N][N];
int dx,dy,tx,ty,n;
int ddx[]={1,0,0,-1};
int ddy[]={0,1,-1,0};
int vis[N][N];
struct node{
int x,y,t;
char b;
};
int bfs()
{
vis[dx][dy] = 1;
queue<node>q;
q.push({dx,dy,0,'r'});
while(q.size())
{
node z = q.front();
q.pop();
if(z.x==tx&&z.y==ty)
{
return z.t;
}
for(int i=0;i<4;i++)
{
int x = z.x+ddx[i];
int y = z.y+ddy[i];
if(!vis[x][y]&&x>0&&y>0&&x<=n&&y<=n&&s[x][y]!=z.b)
{
q.push({x,y,z.t+1,s[x][y]});
vis[x][y] = 1;
}
}
}
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin>>s[i][j];
if(s[i][j]=='A')
{
dx=i;
dy=j;
}
if(s[i][j]=='B')
{
tx=i;
ty=j;
}
}
}
cout<<bfs();
return 0;
}