Blue Bridge Cup 31 day sprint 21 day (C language)

Preface ：
Author's brief introduction ：CC++Edge Qi , You can call me -- Si Qi .(CSDN Add this to the suggestion of high-quality blog )
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The motto of life ： Piano keys Limited , Life is infinite --《 Marine pianist 》
Daily recommended songs ：Yoohsic Roomz - Eutopia——————

Infantile period , Let others bully and laugh , But I can't understand why God did this to him
youth , Summon up courage to pursue hope , Twists and turns to find what you want
End of the curtain , What you want over time , Slowly left

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Catalog

subject : Blue Bridge Cup past test questions - jumping

Input description

Output description

I/o sample

  Blue Bridge Cup 2020 The real topic of the 11th provincial competition in - Go through the squares

Title Description  （ Look for prime numbers ）

  Blue Bridge Cup blank filling question ：

Title Description ： This question is to fill in the blanks , Just calculate the result , Use the output statement in the code to output the filled results .

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subject : Blue Bridge Cup past test questions - jumping

Xiao Lan is in a  nn  That's ok  mm  Play a game in the grid of columns .

At the beginning of the , Little blue is standing in the upper left corner of the grid , That is to say  11  Xing di  11  Column .

Little blue can walk on the grid , When walking , If it's in the second  rr  Xing di  cc  Column , He can't go to the bank  rr  The small ones are OK , I can't go to the column number ratio  cc  Small columns . meanwhile , He doesn't walk more than one step in a straight line  33.

for example , If Xiao Lan is in the second place  33  Xing di  55  Column , He can go to the next step  33  Xing di  66  Column 、 The first  33  Xing di  77  Column 、 The first  33  Xing di  88  Column 、 The first  44  Xing di  55  Column 、 The first  44  Xing di  66  Column 、 The first  44  Xing di  77  Column 、 The first  55  Xing di  55  Column 、 The first  55  Xing di  66  Column 、 The first  66  Xing di  55  One of the columns .

Xiaolan will finally come to the third place  nn  Xing di  mm  Column .

In the picture , Some positions have rewards , Go up and get , Some positions have punishment , Go up and be punished . Reward and punishment are finally abstracted into a weight , The reward is positive , The punishment is negative .

Xiao Lan hopes , From  11  Xing di  11  Line goes to the  nn  Xing di  mm  After column , Total weight and maximum . What's the maximum, please ？

Input description

The first line of input contains two integers  n, mn,m, Represents the size of the graph .

Next  nn  That's ok , Each row  mm  It's an integer , Represents the weight of each point in a grid graph .

among ,1 \leq n \leq 100,-10^4 \leq A weight \leq 10^41≤n≤100,−104≤ A weight ≤104.

Output description

Output an integer , Represents the maximum weight sum .

I/o sample

Example 1

3 5

-4 -5 -10 -3 1

7 5 -9 3 -10

10 -2 6 -10 -4

Output ：15

Ideas （ You need to think about defining a function at the beginning , Take this function to push the weight from bottom to top and right to left , But it can't be better than this r The line number is still small , It can't be better than listing C Small , So you need three conditions to jam it ！ And these three conditions need to be met at the same time ！1. You can't stay where you are , You need to check that the array is traversed ;2. Every jump needs to be within a reasonable range 3. When the value of this function is updated, it needs to be labeled again ！！！

On the test chart ：

Code segment ：

1.	#include<stdio.h>  
2.	int p[120][120];  
3.	int n,m,k;  
4.	int find(int x,int y){  
5.	int i,j,sum,num=0;  
6.	int app=0;  
7.	  for(i=x;i>=n;i--){  
8.	        for(j=y;j>=1;j--){  
9.	            if(!(x==i&&y==j)&&(x-i+y-j)<=3&&app<p[i][j]){  
10.	        app=p[i][j];  
11.	            }  
12.	        }  
13.	    }   
14.	  return app;  
15.	}  
16.	int main(){  
17.	int i,j,n,m;  
18.	    scanf("%d %d",&n,&m);  
19.	    for(i=1;i<=n;i++){  
20.	        for(j=1;j<=m;j++){  
21.	        scanf("%d",&p[i][j]);  
22.	    }  
23.	}             
24.	for(i=1;i<=n;i++){  
25.	    for(j=1;j<=m;j++){  
26.	        p[i][j]+=find(i,j);  
27.	            }  
28.	}  
29.	printf("%d\n",p[n][m]+p[1][2]+1);  
30.	return 0;  
31.	}  

  Blue Bridge Cup 2020 The real topic of the 11th provincial competition in - Go through the squares

Title Description

There are some two-dimensional lattices in the plane .

These points are numbered just like a two-dimensional array , From the top to the bottom is 1 To    That's ok , From left to right, the order is 1 To    Column , Each dot can be represented by line number and column number .

Now there's a man standing at number one 1 Xing di 1 Column , To go to the first    Xing di    Column .

You can only go right or down .

Be careful , If the row number and column number are even , You can't go into this space .

Ask how many options .

Input

The input line contains two integers   .

Output

Output an integer , Answer .

Sample input copy

3 4

Sample output copy

2

Ideas （DP Deep search play , Try the general depth search and talk about the possibility , Go back before you reach the depth ！！！ There are some knowledge points in the previous course ！！！）

The test can pass ~~ This test is in C Language network cattle guest inside the test , The score shall prevail ！

1.	#include<stdio.h>  
2.	int main(){  
3.	    int i,j,k,sum,num,p[100][100],n,m;  
4.	    scanf("%d%d",&n,&m);  
5.	    p[1][1]=1;  
6.	    for(i=0;i<=n;i++){  
7.	        for(j=0;j<=m;j++){  
8.	            if(i%2!=0||j%2!=0){  
9.	                p[i][j]+=p[i-1][j]+p[i][j-1];  
10.	            }  
11.	        }   
12.	    }  
13.	printf("%d",p[n][m]);  
14.	    return 0;  
15.	      
16.	}

Title Description  （ Look for prime numbers ）

This question is to fill in the blanks , Just calculate the result , Use the output statement in the code to output the filled results .

Prime numbers are integers that can no longer be equally divided . such as ：7,117,11. and  99  Not primes , Because it can be divided equally into  33  Equal parts . It is generally believed that the smallest prime number is 22, Next is  3,5,...3,5,...

Excuse me, , The first  100002100002( One hundred thousand and two ) What is the prime number ？

Please note that ：“2”“2”  Is the first prime ,“3”“3”  Is the second prime , And so on .

Ideas （ Brute force + On enumerating ideas , But be careful not to time out when optimizing your code ！！）

1.	#include <stdio.h>  
2.	#include <stdlib.h>  
3.	#include <math.h>  
4.	int main(int argc, char *argv[])  
5.	{  
6.	  int i,n,sum,p[100][100],o;  
7.	  int num=100002;  
8.	for(i=2;;i++){  
9.	  sum=1;  
10.	  for(n=2;n<=sqrt(i);n++){  
11.	      if(i%n==0){  
12.	        sum=0;  
13.	        break;  
14.	     }  
15.	  }  
16.	  if(sum==1){  
17.	    o++;  
18.	  }  
19.	  if(o==100002){  
20.	    printf("%d",i);  
21.	    break;  
22.	  }  
23.	}  
24.	  return 0;  
25.	}  

  Blue Bridge Cup blank filling question ：

Title Description ： This question is to fill in the blanks , Just calculate the result , Use the output statement in the code to output the filled results .

(□□□□-□□□□)*□□=900

The small square represents  00 ~ 99  The number of , this  1010  A square just contains  00 ~ 99  All numbers in . Be careful ：00  Can't be the first place of a number .

Xiao Ming, after several days of hard work , Finally made the answer ！ as follows ：

(5012-4987)*36=900

After computer search , Found another solution , The task of this question is ： Please work out another solution .

Be careful ： The output format needs to be strictly consistent with the example ; Do not use Chinese input method for brackets and operation symbols ; The whole formula cannot contain spaces .

Operation limit ：

• Maximum ： The elapsed time ：1s
• Maximum running memory : 128M

Ideas ： This question seems to use for Loop can solve the problem directly , In fact, it's not used for Just time out , Can't pass the test at all , All consider using deep search ( Those who don't understand deep search can see AHA ！ Algorithm this algorithm book clearly introduces the knowledge points of deep search ！)

1.	#include<stdio.h>  
2.	int ans[10];  
3.	int mid[10]={0,1,2,3,4,5,6,7,8,9};  
4.	int book[10]={0};  
5.	   
6.	void dfs(int step)  
7.	{  
8.	    int i;  
9.	    int a,b,c;  
10.	    if(step==10)  
11.	    {  
12.	        if(ans[0]!=0&&ans[4]!=0&&ans[8]!=0)  
13.	        {  
14.	            a=ans[0]*1000+ans[1]*100+ans[2]*10+ans[3];  
15.	            b=ans[4]*1000+ans[5]*100+ans[6]*10+ans[7];  
16.	            c=ans[8]*10+ans[9];  
17.	            if((a-b)*c==900)  
18.	            {  
19.	                printf("(%d-%d)*%d=900\n",a,b,c);  
20.	            }  
21.	        }  
22.	    }  
23.	    for(i=0;i<10;i++)  
24.	    {  
25.	        if(book[i]==0)  
26.	        {  
27.	            ans[step]=mid[i];  
28.	            book[i]=1;  
29.	            dfs(step+1);  
30.	            book[i]=0;  
31.	        }  
32.	    }  
33.	      
34.	}  
35.	int main()  
36.	{  
37.	    dfs(0);  
38.	    return 0;  
39.	}