$\ \ \ \ \ \ \$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;
}

#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;

}

#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;
}