Here is the mashup link (the problems are from Codeforces' problem set).

#### A. Codeforces' Characters

**Solution**

For every character given check if that character appears inside "codeforces".

**Code**

```
t = int(input())
for _ in range(t):
char = input()
if char in "codeforces":
print("YES")
else:
print("NO")
```

#### B. Add or Subtract

**Solution**

Get the numbers a, b, and c from the console. If the result of subtracting b from a is equal to c, output '-'; otherwise, output '+'.

**Code**

```
t = int(input())
for _ in range(t):
a, b, c = list(map(int, input().split()))
if a - b == c:
print("-")
else:
print("+")
```

#### C. Pneumonoultramicroscopicsilicovolcanoconiosis

**Solution**

For a given word if the length is less than or equal to 10 print the word back, else print "firstletter + length of the word minus 2 + lastword".

**Code**

```
t = int(input())
for _ in range(t):
word = input()
if len(word) > 10:
print(word[0] + str(len(word) - 2) + word[-1])
else:
print(word)
```

#### D. Even Sum

**Hint**

We have to remove at most one number.

**Solution**

First, if the sum of all the numbers is already even, then we do nothing. Otherwise, we remove the smallest odd number. Since, if the sum is odd, we need to remove a number with the same parity to make the sum even. Notice it's always bad to remove more odd numbers, and it does nothing to remove even numbers.

**Code**

```
N = int(input())
nums = list(map(int, input().split()))
total = sum(nums)
if total%2 == 0:
print(total)
else:
min_odd = float('inf')
for num in nums:
if num%2 == 1 and num < min_odd:
min_odd = num
print(total - min_odd)
```

#### E. Million the Painter

**Hint 1**

Consider when there is "??".

**Hint 2**

In how many ways can we paint "C?C"?

**Hint 3**

Consider when there is "?" at the end or beginning.

**Solution**

What causes the answer to be "Yes"? Of course, there cannot be adjacent segments that already have the same colour; but what else? We can figure out that whenever two consecutive question marks appear, there are at least two ways to fill them. But the samples lead us to ponder over cases with one single question mark: a question mark can be coloured in two different colours if it lies on the boundary of the canvas, or is between two adjacent segments of the same colour. Putting it all together, we get a simple but correct solution.

**Code**

```
N = int(input())
s = input()
if ("CC" in s) or ("YY" in s) or ("MM" in s):
print("No")
elif ("??" in s) or ("C?C" in s) or ("Y?Y" in s) or ("M?M" in s) or (s[0] == "?") or (s[-1] == "?"):
print("Yes")
else:
print("No")
```