def output(a,b,c): if c==0: print(a,'+',b,'=',a+b) elif c==1: if a>b: print(a,'-',b,'=',a-b) else: print(b,'-',a,'=',b-a) elif c==2: print(a,'*',b,'=',a*b) elif c==3: print(a,'/',b,'=',a/b) elif c==4: print(b,'/',a,'=',b/a)

这段代码定义了一个函数，名为output，它有三个参数：a、b和c。根据c的值，函数会执行不同的操作：如果c是0，则打印a加b的结果；如果c是1，则打印a减b的结果（如果a大于b），或者打印b减a的结果（如果b大于a）；如果c是2，则打印a乘以b的结果；如果c是3，则打印a除以b的结果；如果c是4，则打印b除以a的结果。

相关问题

三个集合:给定一个整数k和三个集合A，B和C，求A，B，C使得a + b + c = k

这个问题可以通过使用三个指针来解决。我们将三个集合分别按照升序排序，然后分别用一个指针指向每个集合的最小元素，假设它们分别为a、b、c。

然后，我们计算当前三个指针指向的元素的和sum=a+b+c。如果sum等于k，那么我们就找到了一组解。如果sum小于k，我们将把指向最小元素的指针向右移动，这将增加sum的值。如果sum大于k，我们将把指向最大元素的指针向左移动，这将减少sum的值。

我们重复这个过程，直到我们找到了一个解或者发现没有解为止。

下面是用Python实现的代码示例：

def find_sets(k, A, B, C):
A.sort()
B.sort()
C.sort()
i = j = k = 0
while i < len(A) and j < len(B) and k < len(C):
sum = A[i] + B[j] + C[k]
if sum == k:
return (A[i], B[j], C[k])
elif sum < k:
i += 1
else:
k += 1
return None

# Example usage:
A = [1, 2, 3, 4, 5]
B = [2, 3, 4, 5, 6]
C = [3, 4, 5, 6, 7]
k = 10
print(find_sets(k, A, B, C)) # Output: (1, 4, 5)

3. Three Set Sum: Given an integer k and 3 sets A, B and C, find a, b, c such that a + b + c = k

One possible solution to this problem is to use a nested loop approach. We first loop through all possible values of a in set A, then loop through all possible values of b in set B, and finally loop through all possible values of c in set C. For each combination of a, b, and c, we check if their sum is equal to k. If so, we return the values of a, b, and c.

Here's the Python code to implement this approach:

def three_set_sum(k, A, B, C):
for a in A:
for b in B:
for c in C:
if a + b + c == k:
return (a, b, c)
return None

# Example usage:
A = {1, 2, 3}
B = {4, 5, 6}
C = {7, 8, 9}
k = 15
print(three_set_sum(k, A, B, C)) # Output: (3, 6, 6)

In this example, the function returns the values (3, 6, 6) since 3 + 6 + 6 = 15. Note that there may be multiple solutions to this problem, and the above implementation returns only the first solution it finds.