4 条题解

  • 1
    @ 2023-10-8 16:12:31
    #include<stdio.h>
    #include<math.h>
     
    int main()
    {
    	int n,count=0;
    	int a[1001][4]={0},sum[1001]={0};
    	
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++)
    	{
    		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);
    		sum[i]=a[i][1]+a[i][2]+a[i][3];
    	}
    	
    	for(int i=1;i<=n;i++)
    	{
    		for(int j=i+1;j<=n;j++)
    		{
    			if(abs(a[i][1]-a[j][1])<=5 &&
    			   abs(a[i][2]-a[j][2])<=5 &&
    			   abs(a[i][3]-a[j][3])<=5 &&
    			   abs(sum[i]-sum[j])<=10)
    			   count++;
    		}
    	}
    	printf("%d",count);
    	
    	return 0;
    }
    
  • 1
    @ 2023-8-26 20:46:33

           \ \ \ \ \ \ \ 蚌埠住了,凭什么我的Python过不去。

    n = int(input())
    students = []
    ans = 0
    for _ in range(n):
        lst = list(map(int, input().split( )))
        students.append(lst)
    for i in range(n - 1):
        for j in range(i + 1, n):
            if (-5 <= students[i][0] - students[j][0] <= 5) and (-5 <= students[i][1] - students[j][1] <= 5) and (-5 <= students[i][2] - students[j][2] <= 5) and (sum(students[j]) - sum(students[i]) <= 10):
                ans += 1
    print(ans)
    

           \ \ \ \ \ \ \ 但是C++的过了。麻。

    #include<cstdio>
    #include<cmath>
    
    const int MAXN = 1010;
    
    struct Stu{
    	int chinese, math, english;
    }students[MAXN];
    
    int main(){
    	int n, ans;
    	scanf("%d", &n);
    	for(int i = 1; i <= n; i++){
    		scanf("%d %d %d", &students[i].chinese, &students[i].math, &students[i].english);
    	}
    	for(int i = 1; i <= n; i++){
    		for(int j = i + 1; j <= n; j++){
    			if((abs(students[i].chinese - students[j].chinese) <= 5) && (abs(students[i].math - students[j].math) <= 5) && (abs(students[i].english - students[j].english) <= 5) && (abs(students[i].chinese - students[j].chinese + students[i].math - students[j].math + students[i].english - students[j].english) <= 10)){
                    ans ++;
                }
    		}
    	}
        printf("%d", ans);
    	return 0;
    }
    
    • 0
      @ 2025-11-8 0:16:32
      #include <stdio.h>
      struct Students
      {
          int Chain;
          int Mach;
          int English;
          int All;
      };
      int search(struct Students a[],int n)
      {
          int m=0;
          for(int i=0;i<n;i++)
          {
              for(int k=0;k<n;k++)
              {    
                  int A,B,C,D;
                  if(k==i){continue;}
      
                  A=a[i].All-a[k].All;
                  B=a[i].Mach-a[k].Mach;
                  C=a[i].Chain-a[k].Chain;
                  D=a[i].English-a[k].English;
      
                  if(A>=-10&&A<=10)
                  {
                      if(B>=-5&&B<=5)
                      {
                          if(C>=-5&&C<=5)
                          {
                              if(D>=-5&&D<=5)
                              {
                                  m++;
                              }
                          }
                          
                      }
                      
                  }
                  
              }
          }
          return m;
      }
      int main()
      {
          struct Students a[1001];
          int n;
          scanf("%d",&n);
          for(int i=0;i<n;i++)
          {
              scanf("%d %d %d",&a[i].Chain,&a[i].Mach,&a[i].English);
              a[i].All=a[i].Chain+a[i].Mach+a[i].English;
          }
          printf("%d",search(a,n)/2);
          return 0;
      }
      
      • 0
        @ 2024-11-19 19:56:39

        #include<stdio.h>

        #include<math.h>

        typedef struct student{

        int a;

        int b;

        int c;

        }STU;

        int main(){

        int n;

        scanf("%d",&n);

        STU ai[n];

        for(int i=0;i<n;i++){

        scanf("%d %d %d",&ai[i].a,&ai[i].b,&ai[i].c); }

        int find = 0;

        for(int i=0;i<n;i++){

        for(int k=i+1;k<n;k++){

        int sum = (ai[i].a+ai[i].b+ai[i].c)- (ai[k].a+ai[k].b+ai[k].c);

        if(abs(ai[i].a-ai[k].a)<=5&&abs(ai[i].b-ai[k].b)<=5&&abs(ai[i].c-ai[k].c)<=5&&abs(sum)<=10) {

        find++;

        }

        }

        }

        printf("%d",find);

        return 0;

        }

        • 1

        信息

        ID
        98
        时间
        1000ms
        内存
        256MiB
        难度
        6
        标签
        (无)
        递交数
        801
        已通过
        218
        上传者