message stringlengths 2 43.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 853 107k | cluster float64 24 24 | __index_level_0__ int64 1.71k 214k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,545 | 24 | 147,090 |
Tags: greedy, implementation
Correct Solution:
```
#!python3
import sys
def calculate(c1, c2, c3, a1, a2, a3, a4, a5):
c1 -= a1
if c1 < 0:
return 'NO'
c2 -= a2
if c2 < 0:
return 'NO'
c3 -= a3
if c3 < 0:
return 'NO'
a4 -= min(a4, c1)
a5 -= min(a5, c2)
if c3 < a4 + a5:
return 'NO'
return 'YES'
test_count = int(sys.stdin.readline())
for _ in range(test_count):
c1, c2, c3 = map(int, sys.stdin.readline().strip().split(' '))
a1, a2, a3, a4, a5 = map(int, sys.stdin.readline().strip().split(' '))
answer = calculate(c1, c2, c3, a1, a2, a3, a4, a5)
sys.stdout.write(f'{answer}\n')
``` | output | 1 | 73,545 | 24 | 147,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,546 | 24 | 147,092 |
Tags: greedy, implementation
Correct Solution:
```
for t in range(int(input())):c=list(map(int,input().split()));a=list(map(int,input().split()));print('YES') if c[0]>=a[0] and c[1]>=a[1] and c[2]>=a[2] and c[0]+c[2]>=a[0]+a[2]+a[3] and c[1]+c[2]>=a[1]+a[2]+a[4] and sum(c)>=sum(a) else print('NO')
``` | output | 1 | 73,546 | 24 | 147,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,547 | 24 | 147,094 |
Tags: greedy, implementation
Correct Solution:
```
cases = int(input())
while cases:
cases -= 1
c1, c2, c3 = map(int, input().split())
a1, a2, a3, a4, a5 = map(int, input().split())
if a1 > c1 or a2 > c2 or a3 > c3 or a1+a3+a4 > c1 + c3 or a2 + a3 + a5 > c2 + c3 or a1+a2+a3+a4+a5 > c1+c2+c3:
print("NO")
else:
print("YES")
``` | output | 1 | 73,547 | 24 | 147,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,548 | 24 | 147,096 |
Tags: greedy, implementation
Correct Solution:
```
import sys
for _ in range(int(sys.stdin.readline().strip())):
c=list(map(int,sys.stdin.readline().strip().split(" ")))
a=list(map(int,sys.stdin.readline().strip().split(" ")))
saab=True
for i in range(len(c)):
c[i]-=a[i]
for i in c:
if i<0:
saab=False
break
if a[3]>c[0]:
a[3]-=c[0]
c[2]-=a[3]
if a[4]>c[1]:
a[4]-=c[1]
c[2]-=a[4]
if c[2]<0:
saab=False
if saab:
print("YES")
else:
print("No")
``` | output | 1 | 73,548 | 24 | 147,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,549 | 24 | 147,098 |
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for i in range(t):
b1, b2, b3 = map(int, input().split())
t1, t2, t3, t4, t5 = map(int, input().split())
if b1>=t1 and b2>=t2 and b3>=t3 and b1+b3>=t1+t3+t4 and b2+b3>=t2+t3+t5 and b1+b2+b3>=t1+t2+t3+t4+t5:
print('YES')
else:
print('NO')
``` | output | 1 | 73,549 | 24 | 147,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,550 | 24 | 147,100 |
Tags: greedy, implementation
Correct Solution:
```
# e068ba04541cf7ebd939e9410394a3cf36a875e9525d69017060edfc1d7eb62c
for _ in range(int(input())):
c = list(map(int, input().split()))
a = list(map(int, input().split()))
if c[0] >= a[0] and c[1] >= a[1] and c[2] >= a[2]:
c = [c[i] - a[i] for i in range(3)]
a[3] -= c[0]
a[4] -= c[1]
if max(0,a[3]) + max(0,a[4]) <= c[2]:
print("YES")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 73,550 | 24 | 147,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. | instruction | 0 | 73,551 | 24 | 147,102 |
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
for _ in range(t):
c=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
for i in range(3):
c[i]=c[i]-a[i]
if c[0]<0 or c[1]<0 or c[2]<0:
print('No')
else:
a[3]=max(0,a[3]-c[0])
a[4]=max(0,a[4]-c[1])
if c[2]>=a[3]+a[4]:
print('Yes')
else:
print('No')
``` | output | 1 | 73,551 | 24 | 147,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
import string
ORDA = 97 # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=2):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
new_number = 0
while number > 0:
new_number += number % base
number //= base
return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s):
for i in range(len(s) // 2 + 1):
if s[i] != s[-i - 1]:
return False
return True
for _ in range(ii()):
c = li()
a = li()
if c[0] < a[0] or c[1] < a[1] or c[2] < a[2]:
print('NO')
else:
c[0] -= a[0]
a[0] = 0
c[1] -= a[1]
a[1] = 0
c[2] -= a[2]
a[2] = 0
remains = min(c[0], a[3])
c[0] -= remains
a[3] -= remains
remains = min(c[1], a[4])
c[1] -= remains
a[4] -= remains
if sum(a[3:]) > c[2]:
print('NO')
else:
print('YES')
``` | instruction | 0 | 73,552 | 24 | 147,104 |
Yes | output | 1 | 73,552 | 24 | 147,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
results = []
num_test = int(input())
for i in range(num_test):
containers = [int(x) for x in input().split()]
trash = [int(x) for x in input().split()]
paper_left = max(trash[0] + trash[3] - containers[0], 0)
plastic_left = max(trash[1] + trash[4] - containers[1], 0)
if (containers[0] < trash[0]) or (containers[1] < trash[1]) or (containers[2] < trash[2]):
results.append("NO")
elif (paper_left + plastic_left) <= containers[2] - trash[2]:
results.append("YES")
else:
results.append("NO")
[print(x) for x in results]
``` | instruction | 0 | 73,553 | 24 | 147,106 |
Yes | output | 1 | 73,553 | 24 | 147,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
#import sys
#import math
#sys.stdout=open("python/output.txt","w")
#sys.stdin=open("python/input.txt","r")
t=int(input())
for i in range(t):
c1,c2,c3=map(int,input().split())
a1,a2,a3,a4,a5=map(int,input().split())
if a1>c1 or a2>c2 or a3>c3:
print("NO")
else:
od=c1-a1
td=c2-a2
a4-=od
a5-=td
if a4>=0 and a5>=0:
a3+=a4+a5
if a3<=c3:
print("YES")
else:
print("NO")
elif a4>=0:
a3+=a4
if a3<=c3:
print("YES")
else:
print("NO")
else:
a3+=a5
if a3<=c3:
print("YES")
else:
print("NO")
``` | instruction | 0 | 73,554 | 24 | 147,108 |
Yes | output | 1 | 73,554 | 24 | 147,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
for _ in range(int(input())):
c=list(map(int, input().split()))
a=list(map(int, input().split()))
for i in range(3):
c[i]-=a[i]
a[i]=0
f=0
if c[0]<0 or c[1]<0:
f=1
if f:
print('NO')
else:
k=min(c[0], a[3])
c[0]-=k
a[3]-=k
k=min(c[1], a[4])
c[1]-=k
a[4]-=k
k=min(c[2], a[3])
c[2] -= k
a[3] -= k
k = min(c[2], a[4])
c[2] -= k
a[4] -= k
for i in a:
if i>0:
f=1
if f:
print('NO')
else:
print('YES')
``` | instruction | 0 | 73,555 | 24 | 147,110 |
Yes | output | 1 | 73,555 | 24 | 147,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
T = int(input())
for t in range(0, T):
string1 = list(map(int, input().split(" ")))
c1 = string1[0]
c2 = string1[1]
c3 = string1[2]
string2 = list(map(int, input().split(" ")))
a1 = string2[0]
a2 = string2[1]
a3 = string2[2]
a4 = string2[3]
a5 = string2[4]
p1 = a1 + a4 - c1
p2 = a2 + a5 - c2
p3 = c3 - a3
if a1 > c1:
print("NO")
elif a2 > c2:
print("NO")
elif a3 > c3:
print("NO")
elif p1<0:
if p2<=p3:
print("YES")
else:
print("NO")
elif p2< 0:
if p1 < p3:
print("YES")
else:
print("NO")
elif p3< 0:
print("NO")
elif p1 + p2 <= p3:
print("YES")
else:
print("NO")
``` | instruction | 0 | 73,556 | 24 | 147,112 |
No | output | 1 | 73,556 | 24 | 147,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
import sys
import math
def scan(input_type='int'):
if input_type == 'int':
return list(map(int, sys.stdin.readline().strip().split()))
else:
return list(map(str, sys.stdin.readline().strip()))
def solution():
for _ in range(int(input())):
c = scan()
a = scan()
c[0] -= a[0]
c[1] -= a[1]
c[2] -= a[2]
if c[0] < 0 or c[1] < 0 or c[2] < 0:
print('No')
continue
if a[3] >= c[0]:
a[3] -= c[0]
c[0] = 0
if a[4] >= c[1]:
a[4] -= c[1]
c[1] = 0
if a[3] + a[4] <= c[2]:
print('Yes')
else:
print('No')
if __name__ == '__main__':
solution()
``` | instruction | 0 | 73,557 | 24 | 147,114 |
No | output | 1 | 73,557 | 24 | 147,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
for i in range(int(input())):
c1, c2, c3 = map(int, input().split())
a1, a2, a3, a4, a5 = map(int, input().split())
p = a1+a4
q = a2+a5
r = abs(c1-p)
t = abs(c2-q)
if c1 >= a1 and c2 >= a2 and c3 >= a3 and c3 >= a3+r+t:
print("YES")
else:
print("NO")
``` | instruction | 0 | 73,558 | 24 | 147,116 |
No | output | 1 | 73,558 | 24 | 147,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.
Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one β all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.
Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic β so Monocarp has to put them into the third container (he has a_3 such items).
Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item β for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items β into the third one.
Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one β no more than c_2 items, and the third one β no more than c_3 items?
Input
The first line contains one integer t (1 β€ t β€ 3 β
10^4) β the number of test cases.
Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 β€ c_1, c_2, c_3 β€ 10^8) β the capacities of the containers.
The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 β€ a_i β€ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).
Output
For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47
Output
YES
YES
NO
YES
YES
NO
YES
Note
Explanations for the example test cases:
1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container;
3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items;
4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container;
5. there is no waste at all, so all containers can be left empty;
6. there's no answer since it's impossible to put a paper item into the third container;
7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container.
Submitted Solution:
```
for _ in range(int(input())):
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = 0
if (b[0] + b[3]) <=a[0]:
c +=1
else:
if (a[0]-b[0])<=b[3]:
c +=1
else:
print('NO')
if (b[1]+b[4]) <=a[1]:
c+=1
else:
if (a[1]-b[1])<=b[4]:
c+=1
else:
print('NO')
if ((b[2]+b[0]-a[0] + b[3]) <=a[2]) and ((b[2] -a[1]+ b[1]+b[4]) <= a[2]):
c+=1
else:
print('NO')
if c ==3:
print('YES')
``` | instruction | 0 | 73,559 | 24 | 147,118 |
No | output | 1 | 73,559 | 24 | 147,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has a sequence, consisting of n non-negative integers: a1, a2, ..., an.
Let's define function f(l, r) (l, r are integer, 1 β€ l β€ r β€ n) for sequence a as an operation of bitwise OR of all the sequence elements with indexes from l to r. Formally: f(l, r) = al | al + 1 | ... | ar.
Polycarpus took a piece of paper and wrote out the values of function f(l, r) for all l, r (l, r are integer, 1 β€ l β€ r β€ n). Now he wants to know, how many distinct values he's got in the end.
Help Polycarpus, count the number of distinct values of function f(l, r) for the given sequence a.
Expression x | y means applying the operation of bitwise OR to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is marked as "|", in Pascal β as "or".
Input
The first line contains integer n (1 β€ n β€ 105) β the number of elements of sequence a. The second line contains n space-separated integers a1, a2, ..., an (0 β€ ai β€ 106) β the elements of sequence a.
Output
Print a single integer β the number of distinct values of function f(l, r) for the given sequence a.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Examples
Input
3
1 2 0
Output
4
Input
10
1 2 3 4 5 6 1 2 9 10
Output
11
Note
In the first test case Polycarpus will have 6 numbers written on the paper: f(1, 1) = 1, f(1, 2) = 3, f(1, 3) = 3, f(2, 2) = 2, f(2, 3) = 2, f(3, 3) = 0. There are exactly 4 distinct numbers among them: 0, 1, 2, 3. | instruction | 0 | 75,335 | 24 | 150,670 |
Tags: bitmasks
Correct Solution:
```
#CF Round 150. Div II Prob. C - The Brand New Function
import sys
In = sys.stdin
n = int(In.readline().strip())
arr = [int(x) for x in In.readline().split()]
res, m = set(), set()
for i in range(n):
s = set()
s.add(arr[i])
s.update([x | arr[i] for x in m])
m = s
res.update(s)
print(len(res))
``` | output | 1 | 75,335 | 24 | 150,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus has a sequence, consisting of n non-negative integers: a1, a2, ..., an.
Let's define function f(l, r) (l, r are integer, 1 β€ l β€ r β€ n) for sequence a as an operation of bitwise OR of all the sequence elements with indexes from l to r. Formally: f(l, r) = al | al + 1 | ... | ar.
Polycarpus took a piece of paper and wrote out the values of function f(l, r) for all l, r (l, r are integer, 1 β€ l β€ r β€ n). Now he wants to know, how many distinct values he's got in the end.
Help Polycarpus, count the number of distinct values of function f(l, r) for the given sequence a.
Expression x | y means applying the operation of bitwise OR to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is marked as "|", in Pascal β as "or".
Input
The first line contains integer n (1 β€ n β€ 105) β the number of elements of sequence a. The second line contains n space-separated integers a1, a2, ..., an (0 β€ ai β€ 106) β the elements of sequence a.
Output
Print a single integer β the number of distinct values of function f(l, r) for the given sequence a.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use cin, cout streams or the %I64d specifier.
Examples
Input
3
1 2 0
Output
4
Input
10
1 2 3 4 5 6 1 2 9 10
Output
11
Note
In the first test case Polycarpus will have 6 numbers written on the paper: f(1, 1) = 1, f(1, 2) = 3, f(1, 3) = 3, f(2, 2) = 2, f(2, 3) = 2, f(3, 3) = 0. There are exactly 4 distinct numbers among them: 0, 1, 2, 3. | instruction | 0 | 75,336 | 24 | 150,672 |
Tags: bitmasks
Correct Solution:
```
n, p, q = input(), set(), set()
for i in map(int, input().split()):
q = set(i | j for j in q)
q.add(i)
p.update(q)
print(len(p))
# Made By Mostafa_Khaled
``` | output | 1 | 75,336 | 24 | 150,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,111 | 24 | 152,222 |
Tags: greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
sum = 0
for i in a:
sum += i
avg = sum/n
if avg % 1 != 0:
print(-1)
else:
avg = int(avg)
count = 0
for i in a:
if i > avg:
count += 1
print(count)
``` | output | 1 | 76,111 | 24 | 152,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,112 | 24 | 152,224 |
Tags: greedy, math
Correct Solution:
```
for i in range(int(input())):
n=int(input())
l=list(map(int,input().strip().split()))
a=sum(l)/n
l.sort()
c=0
if(a.is_integer()==False):
print(-1)
else:
for i in range(len(l)):
if(l[i]>a):
c+=1
print(c)
``` | output | 1 | 76,112 | 24 | 152,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,113 | 24 | 152,226 |
Tags: greedy, math
Correct Solution:
```
import bisect
from collections import Counter
def main(a):
s = sum(a)
n = len(a)
if s % n != 0:
return -1
m = s // n
ct = sum(1 if x > m else 0 for x in a)
return ct
if __name__ == '__main__':
t = int(input())
outputs = []
for _ in range(t):
n = int(input())
# s = input()
arr = list(map(int, input().split(" ")))
output = main(arr)
outputs.append(output)
for o in outputs:
print(o)
``` | output | 1 | 76,113 | 24 | 152,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,114 | 24 | 152,228 |
Tags: greedy, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
m=sum(a)/n
k=0
if m!=int(m):
print(-1)
continue
else:
for j in range(n):
if a[j]>m:
k=k+1
print(k)
``` | output | 1 | 76,114 | 24 | 152,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,115 | 24 | 152,230 |
Tags: greedy, math
Correct Solution:
```
t = int(input())
def solve():
n = int(input())
arr = list(map(int, input().split()))
s = sum(arr)
if(s % n != 0):
print(-1)
return
same = s//n
offset = 0
for i in arr:
offset += abs(i - same)
arr.sort()
arr.reverse()
if(offset == 0):
print(0)
return
count = 0
for i in arr:
offset -= 2 * (i - same)
count += 1
if(offset <= 0):
break
print(count)
for i in range(t):
solve()
``` | output | 1 | 76,115 | 24 | 152,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,116 | 24 | 152,232 |
Tags: greedy, math
Correct Solution:
```
for i in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
S = sum(a)
if S % n == 0:
asn = 0
for el in a:
if el > S // n:
asn += 1
print(asn)
else:
print(-1)
``` | output | 1 | 76,116 | 24 | 152,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,117 | 24 | 152,234 |
Tags: greedy, math
Correct Solution:
```
t = int(input())
for jj in range(t):
nn = int(input())
candle = list(map(int, input().split()))
if sum(candle)%nn != 0:
print(-1)
continue
else:
ideal = sum(candle)//nn
count = sum(i > ideal for i in candle)
print(count)
``` | output | 1 | 76,117 | 24 | 152,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0 | instruction | 0 | 76,118 | 24 | 152,236 |
Tags: greedy, math
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
if s%n==0:
cd=s//n
ans=0
for i in a:
if i>cd:
ans+=1
print(ans)
else:
print(-1)
``` | output | 1 | 76,118 | 24 | 152,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
for i in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
if(sum(l)%n):
print("-1")
else:
k = sum(l)/n
ans=0
for i in l:
if(i>k):
ans+=1
print(ans)
``` | instruction | 0 | 76,119 | 24 | 152,238 |
Yes | output | 1 | 76,119 | 24 | 152,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
def GetList(): return list(map(int, input().split()))
def solve():
t = int(input())
for num in range(t):
n = int(input())
arr = GetList()
arr = sorted(arr)
if sum(arr)%n!=0:
print("-1")
else:
ans = sum(arr)//n
i = 0
for candies in arr:
if candies>ans:
break
i+=1
print(n-i)
solve()
``` | instruction | 0 | 76,120 | 24 | 152,240 |
Yes | output | 1 | 76,120 | 24 | 152,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
t=int(input())
for l in range(t):
n=int(input())
a=[int(i) for i in input().split(" ")]
s=sum(a)
ans=0
if(s%n!=0):
print(-1)
else:
for i in a:
if(i>s/n):
ans+=1
print(ans)
``` | instruction | 0 | 76,121 | 24 | 152,242 |
Yes | output | 1 | 76,121 | 24 | 152,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
import sys
import math
from math import factorial, inf, gcd, sqrt
from heapq import *
from functools import *
from itertools import *
from collections import *
from typing import *
from bisect import *
import random
sys.setrecursionlimit(10**5)
def rarray():
return [int(i) for i in input().split()]
t = 1
t = int(input())
for ii in range(t):
n = int(input())
a = rarray()
a.sort()
s = sum(a)
if s % n != 0:
print(-1)
continue
k = s // n
print(n - bisect_right(a, k))
``` | instruction | 0 | 76,122 | 24 | 152,244 |
Yes | output | 1 | 76,122 | 24 | 152,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split(' ')]
if sum(a) % len(a) != 0:
print(-1)
else:
avg = sum(a) // len(a)
lw, h = 0, 0
for e in a:
if e < avg:
lw += 1
elif e > avg:
h += 1
print(max(lw, h))
``` | instruction | 0 | 76,123 | 24 | 152,246 |
No | output | 1 | 76,123 | 24 | 152,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
from math import floor, sqrt, gcd
def vmmv(n):
xx=[]
while n%2==0:
n//=2
xx.append(2)
for i in range(3,floor(sqrt(n))+1,2) :
while n%i==0:
xx.append(i)
n//=i
if n>1:
xx.append(n)
return len(xx)
for _ in range(int(input())):
n=int(input())
a=[int(X) for X in input().split()]
if sum(a)%n!=0:
print(-1)
else:
an=0
an12=0
for i in a:
if i>sum(a)//n:an+=1
else:an12+=1
print(min(an,an12))
``` | instruction | 0 | 76,124 | 24 | 152,248 |
No | output | 1 | 76,124 | 24 | 152,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
from sys import stdin
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
if n == 1 or len(set(a)) == 1:
print(0)
continue
s = sum(a)
if s % n != 0:
print(-1)
else:
v = s // n
c = a.count(v)
print(n - c - 1)
``` | instruction | 0 | 76,125 | 24 | 152,250 |
No | output | 1 | 76,125 | 24 | 152,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once:
* Polycarp chooses k (0 β€ k β€ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, β¦, i_k);
* Polycarp distributes their a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + β¦ + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process.
Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.
For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies:
* Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3).
Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.
For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^4).
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output:
* the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way;
* "-1" if no such value k exists.
Example
Input
5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1
Output
2
1
-1
0
0
Submitted Solution:
```
t=int(input())
while(t>0):
n=int(input())
a=[int(i) for i in input().split()]
if((sum(a)%n)!=0):
print(-1)
else:
c=0
req=sum(a)//n
for i in a:
if(i!=req):
c+=1
if(c!=0):
print(c-1)
else:
print(0)
t-=1
``` | instruction | 0 | 76,126 | 24 | 152,252 |
No | output | 1 | 76,126 | 24 | 152,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,235 | 24 | 152,470 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
__author__ = 'zhan'
[a1, b1] = [int(i) for i in input().split()]
[a2, b2] = [int(i) for i in input().split()]
q1 = [[a1, b1, 0]]
q2 = [[a2, b2, 0]]
tested_total1 = []
tested_total2 = []
def equal(t, q):
lo = 0
hi = len(q)
while True:
if lo >= hi:
return False
m = (lo + hi) // 2
p = q[m]
temp = p[0] * p[1]
if t == temp:
return [p[0], p[1], p[2]]
if t < temp:
lo = m + 1
elif t > temp:
hi = m
while True:
if len(q1) > 0 and len(q2) > 0:
total1 = q1[0][0] * q1[0][1]
total2 = q2[0][0] * q2[0][1]
if total1 > total2:
ans = equal(total1, q2)
if ans:
print(str(ans[2] + q1[0][2]) + "\n" + str(q1[0][0]) + " " + str(q1[0][1]) + "\n" + str(ans[0]) + " " + str(ans[1]))
else:
if not(q1[0][0] & 1):
tt = [q1[0][0] // 2, q1[0][1], q1[0][2] + 1]
if not tt[0]*tt[1] in tested_total1:
q1.append(tt)
tested_total1.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q2)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if q1[0][0] % 3 == 0:
tt = [q1[0][0] // 3 * 2, q1[0][1], q1[0][2] + 1]
if not tt[0]*tt[1] in tested_total1:
q1.append(tt)
tested_total1.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q2)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if not(q1[0][1] & 1):
tt = [q1[0][0], q1[0][1] // 2, q1[0][2] + 1]
if not tt[0]*tt[1] in tested_total1:
q1.append(tt)
tested_total1.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q2)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if q1[0][1] % 3 == 0:
tt = [q1[0][0], q1[0][1] // 3 * 2, q1[0][2] + 1]
if not tt[0]*tt[1] in tested_total1:
q1.append(tt)
tested_total1.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q2)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
q1.pop(0)
q1.sort(key=lambda x: x[0]*x[1], reverse=True)
elif total1 < total2:
ans = equal(total2, q1)
if ans:
print(str(ans[2] + q2[0][2]) + "\n" + str(ans[0]) + " " + str(ans[1]) + "\n" + str(q2[0][0]) + " " + str(q2[0][1]))
break
else:
if not(q2[0][0] & 1):
tt = [q2[0][0] // 2, q2[0][1], q2[0][2] + 1]
if not tt[0]*tt[1] in tested_total2:
q2.append(tt)
tested_total2.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q1)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if q2[0][0] % 3 == 0:
tt = [q2[0][0] // 3 * 2, q2[0][1], q2[0][2] + 1]
if not tt[0]*tt[1] in tested_total2:
q2.append(tt)
tested_total2.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q1)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if not(q2[0][1] & 1):
tt = [q2[0][0], q2[0][1] // 2, q2[0][2] + 1]
if not tt[0]*tt[1] in tested_total2:
q2.append(tt)
tested_total2.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q1)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
if q2[0][1] % 3 == 0:
tt = [q2[0][0], q2[0][1] // 3 * 2, q2[0][2] + 1]
if not tt[0]*tt[1] in tested_total2:
q2.append(tt)
tested_total2.append(tt[0]*tt[1])
#an = equal(tt[0]*tt[1], q1)
#if ans:
# print(str(an[2] + tt[2]) + "\n" + str(tt[0]) + " " + str(tt[1]) + "\n" + str(an[0]) + " " + str(an[1]))
q2.pop(0)
q2.sort(key=lambda x: x[0]*x[1], reverse=True)
else:
print(str(q1[0][2] + q2[0][2]) + "\n" + str(q1[0][0]) + " " + str(q1[0][1]) + "\n" + str(q2[0][0]) + " " + str(q2[0][1]))
break
else:
print(-1)
break
``` | output | 1 | 76,235 | 24 | 152,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,236 | 24 | 152,472 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
__author__ = 'zhan'
[a1, b1] = [int(i) for i in input().split()]
[a2, b2] = [int(i) for i in input().split()]
q = [[[a1, b1, 0]], [[a2, b2, 0]]]
total = [0, 0]
tested = [[], []]
def expand(i):
if not (q[i][0][0] & 1):
tt = [q[i][0][0] // 2, q[i][0][1], q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if q[i][0][0] % 3 == 0:
tt = [q[i][0][0] // 3 * 2, q[i][0][1], q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if not (q[i][0][1] & 1):
tt = [q[i][0][0], q[i][0][1] // 2, q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if q[i][0][1] % 3 == 0:
tt = [q[i][0][0], q[i][0][1] // 3 * 2, q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
q[i].pop(0)
q[i].sort(key=lambda x: x[0] * x[1], reverse=True)
while True:
if len(q[0]) > 0 and len(q[1]) > 0:
total[0] = q[0][0][0] * q[0][0][1]
total[1] = q[1][0][0] * q[1][0][1]
if total[0] > total[1]:
expand(0)
elif total[0] < total[1]:
expand(1)
else:
print(str(q[0][0][2] + q[1][0][2]) + "\n" + str(q[0][0][0]) + " " + str(q[0][0][1]) + "\n" + str(
q[1][0][0]) + " " + str(q[1][0][1]))
break
else:
print(-1)
break
``` | output | 1 | 76,236 | 24 | 152,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,237 | 24 | 152,474 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
def follow(x, its=0):
yield (x, its)
while x % 2 == 0:
its += 1
x //= 2
yield (x, its)
def segment(x, its=0):
yield from follow(x, its=its)
while x % 3 == 0:
its += 1
x = x // 3 * 2
yield from follow(x, its=its)
def joinseg(a, b):
for v1, c1 in a:
for v2, c2 in b:
yield (v1*v2, (c1+c2, v1, v2))
def best(it, dic):
return min(dic[x] for x in it)
bar1 = dict(joinseg(*map(lambda x: list(segment(int(x))), input().split(' '))))
bar2 = dict(joinseg(*map(lambda x: list(segment(int(x))), input().split(' '))))
keys = bar1.keys() & bar2.keys()
if not keys:
print(-1)
else:
vals, v1, v2 = zip(best(keys, bar1), best(keys, bar2))
b1v, b2v = zip(v1, v2)
print(sum(vals))
print(*b1v)
print(*b2v)
``` | output | 1 | 76,237 | 24 | 152,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,238 | 24 | 152,476 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
a,b=map(int,input().split())
c,d=map(int,input().split())
e=a*b
f=c*d
n=0
while e%2==0:e=e//2
while e%3==0:e=e//3
while f%2==0:f=f//2
while f%3==0:f=f//3
if e!=f:print("-1")
else:
i=0
j=0
e=a*b
f=c*d
while e%3==0:
e=e//3
i+=1
while f%3==0:
f=f//3
j+=1
k=i-j
if k>0:
for i in range(k):
n+=1
if a%3==0:a=a*2//3
else:b=b*2//3
else:
for i in range(0-k):
n+=1
if c%3==0:c=c*2//3
else:d=d*2//3
e=a*b
f=c*d
i=0
j=0
while e%2==0:
e=e//2
i+=1
while f%2==0:
f=f//2
j+=1
k=i-j
if k>0:
for i in range(k):
n+=1
if a%2==0:a=a//2
else:b=b//2
else:
for i in range(0-k):
n+=1
if c%2==0:c=c//2
else:d=d//2
print(n)
print(a,b)
print(c,d)
``` | output | 1 | 76,238 | 24 | 152,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,239 | 24 | 152,478 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
def decomp(a):
cnt2 = 0
while a%2==0:
a = a//2
cnt2 += 1
cnt3 = 0
while a%3==0:
a = a//3
cnt3 += 1
return a,cnt2,cnt3
def cut(a,b,d,p):
while d>0:
if a%p==0:
a = (p-1)*a//p
d = d-1
elif b%p==0:
b = (p-1)*b//p
d = d-1
return a,b
a1,b1 = [int(s) for s in input().split()]
a2,b2 = [int(s) for s in input().split()]
u1,n2a1,n3a1 = decomp(a1)
v1,n2b1,n3b1 = decomp(b1)
u2,n2a2,n3a2 = decomp(a2)
v2,n2b2,n3b2 = decomp(b2)
##print(u1,v1,u1*v1)
##print(u2,v2,u2*v2)
if u1*v1!= u2*v2:
print(-1)
else:
n = n2a1+n2b1
m = n3a1+n3b1
x = n2a2+n2b2
y = n3a2+n3b2
## print(n,m,x,y)
d3 = abs(m-y)
if m>y:
n += d3
a1,b1 = cut(a1,b1,d3,3)
## print(1,a1,b1)
else:
x += d3
a2,b2 = cut(a2,b2,d3,3)
## print(2,a2,b2)
d2 = abs(n-x)
if n>x:
a1,b1 = cut(a1,b1,d2,2)
## print(1,a1,b1)
else:
a2,b2 = cut(a2,b2,d2,2)
## print(2,a2,b2)
m = d2+d3
print(m)
print(a1,b1)
print(a2,b2)
``` | output | 1 | 76,239 | 24 | 152,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,240 | 24 | 152,480 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
a10 = a1
b10 = b1
a20 = a2
b20 = b2
a_divs = list()
b_divs = list()
div = 2
while a1 > 1 and div < a1 ** 0.5 + 1:
while a1 % div == 0:
a_divs.append(div)
a1 //= div
div += 1
if a1 > 1:
a_divs.append(a1)
div = 2
while b1 > 1 and div < b1 ** 0.5 + 1:
while b1 % div == 0:
a_divs.append(div)
b1 //= div
div += 1
if b1 > 1:
a_divs.append(b1)
div = 2
while a2 > 1 and div < a2 ** 0.5 + 1:
while a2 % div == 0:
b_divs.append(div)
a2 //= div
div += 1
if a2 > 1:
b_divs.append(a2)
div = 2
while b2 > 1 and div < b2 ** 0.5 + 1:
while b2 % div == 0:
b_divs.append(div)
b2 //= div
div += 1
if b2 > 1:
b_divs.append(b2)
a1 = a10
b1 = b10
a2 = a20
b2 = b20
a_divs.sort()
b_divs.sort()
na = len(a_divs)
nb = len(b_divs)
a = 1
while na > 0 and a_divs[-1] > 3:
a *= a_divs[-1]
a_divs = a_divs[:-1]
na -= 1
b = 1
while nb > 0 and b_divs[-1] > 3:
b *= b_divs[-1]
b_divs = b_divs[:-1]
nb -= 1
if a != b:
print(-1)
else:
ans = 0
a_3 = a_divs.count(3)
a_2 = a_divs.count(2)
b_3 = b_divs.count(3)
b_2 = b_divs.count(2)
if a_3 > b_3:
for i in range(a_3 - b_3):
a_divs[a_divs.index(3)] = 2
a_3 -= 1
a_2 += 1
ans += 1
if a1 % 3 == 0:
a1 //= 3
a1 *= 2
else:
b1 //= 3
b1 *= 2
else:
for i in range(b_3 - a_3):
b_divs[b_divs.index(3)] = 2
b_3 -= 1
b_2 += 1
ans += 1
if a2 % 3 == 0:
a2 //= 3
a2 *= 2
else:
b2 //= 3
b2 *= 2
if a_2 > b_2:
for i in range(a_2 - b_2):
a_2 -= 1
ans += 1
if a1 % 2 == 0:
a1 //= 2
else:
b1 //= 2
else:
for i in range(b_2 - a_2):
b_2 -= 1
ans += 1
if a2 % 2 == 0:
a2 //= 2
else:
b2 //= 2
print(ans)
print(a1, b1)
print(a2, b2)
``` | output | 1 | 76,240 | 24 | 152,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,241 | 24 | 152,482 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
f = lambda: map(int, input().split())
a, b = f()
c, d = f()
def g(p, k):
s = 1
while k % p ** s == 0: s += 1
return s - 1
a3, b3, c3, d3 = g(3, a), g(3, b), g(3, c), g(3, d)
a2, b2, c2, d2 = g(2, a), g(2, b), g(2, c), g(2, d)
ab3, cd3 = a3 + b3, c3 + d3
ab2, cd2 = a2 + b2, c2 + d2
ab = a * b * pow(2, cd2) * pow(3, cd3)
cd = c * d * pow(2, ab2) * pow(3, ab3)
if ab != cd:
print(-1)
exit()
k, s2, s3 = 1e9, 0, 0
for t3 in range(min(ab3, cd3) + 1):
k3 = ab3 + cd3 - 2 * t3
for t2 in range(min(ab2 + ab3, cd2 + cd3) - t3 + 1):
k2 = k3 + ab2 + cd2 - 2 * t2
if k2 + k3 < k:
k = k2 + k3
s2, s3 = t2, t3
t3 = ab3 - s3
while t3 and a % 3 == 0:
a = 2 * a // 3
t3 -= 1
while t3 and b % 3 == 0:
b = 2 * b // 3
t3 -= 1
t2 = ab3 - s3 + ab2 - s2
while t2 and a % 2 == 0:
a = a // 2
t2 -= 1
while t2 and b % 2 == 0:
b = b // 2
t2 -= 1
t3 = cd3 - s3
while t3 and c % 3 == 0:
c = 2 * c // 3
t3 -= 1
while t3 and d % 3 == 0:
d = 2 * d // 3
t3 -= 1
t2 = cd3 - s3 + cd2 - s2
while t2 and c % 2 == 0:
c = c // 2
t2 -= 1
while t2 and d % 2 == 0:
d = d // 2
t2 -= 1
print(k)
print(a, b)
print(c, d)
``` | output | 1 | 76,241 | 24 | 152,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1 | instruction | 0 | 76,242 | 24 | 152,484 |
Tags: brute force, dfs and similar, math, meet-in-the-middle, number theory
Correct Solution:
```
#fin = open("input.txt")
#a1, b1 = map(int, fin.readline().split())
#a2, b2 = map(int, fin.readline().split())
a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
F, S = [a1, b1], [a2, b2]
A = dict()
B = dict()
A[2] = A[3] = B[2] = B[3] = 0
i = 2
while i ** 2 <= a1:
if a1 % i == 0:
A[i] = 0
while a1 % i == 0:
A[i] += 1
a1 //= i
i += 1
if a1 > 1:
if not a1 in A:
A[a1] = 0
A[a1] += 1
i = 2
while i ** 2 <= b1:
if b1 % i == 0:
if not i in A:
A[i] = 0
while b1 % i == 0:
A[i] += 1
b1 //= i
i += 1
if b1 > 1:
if not b1 in A:
A[b1] = 0
A[b1] += 1
i = 2
while i ** 2 <= a2:
if a2 % i == 0:
B[i] = 0
while a2 % i == 0:
B[i] += 1
a2 //= i
i += 1
if a2 > 1:
if not a2 in B:
B[a2] = 0
B[a2] += 1
i = 2
while i ** 2 <= b2:
if b2 % i == 0:
if not i in B:
B[i] = 0
while b2 % i == 0:
B[i] += 1
b2 //= i
i += 1
if b2 > 1:
if not b2 in B:
B[b2] = 0
B[b2] += 1
C1 = sorted([i for i in A.keys() if not i in {2, 3}])
C2 = sorted([i for i in B.keys() if not i in {2, 3}])
if C1 != C2:
print(-1)
else:
flag = True
for i in C1:
if (A[i] != B[i]):
flag = False
if not flag:
print(-1)
else:
Min = 0
x = A[3] - B[3]
Min += abs(x)
if x >= 0:
A[2] += x
while x > 0 and F[0] % 3 == 0:
F[0] //= 3
F[0] *= 2
x -= 1
while x > 0 and F[1] % 3 == 0:
F[1] //= 3
F[1] *= 2
x -= 1
else:
B[2] -= x
while x < 0 and S[0] % 3 == 0:
S[0] //= 3
S[0] *= 2
x += 1
while x < 0 and S[1] % 3 == 0:
S[1] //= 3
S[1] *= 2
x += 1
if x != 0:
flag = False
x = A[2] - B[2]
Min += abs(x)
if x >= 0:
while x > 0 and F[0] % 2 == 0:
F[0] //= 2
x -= 1
while x > 0 and F[1] % 2 == 0:
F[1] //= 2
x -= 1
else:
while x < 0 and S[0] % 2 == 0:
S[0] //= 2
x += 1
while x < 0 and S[1] % 2 == 0:
S[1] //= 2
x += 1
if x != 0:
flag = False
if flag:
print(Min)
print(*F)
print(*S)
else:
print(-1)
``` | output | 1 | 76,242 | 24 | 152,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
f = lambda: map(int, input().split())
a, b = f()
c, d = f()
g = lambda k: 0 if k % p else 1 + g(k // p)
k = 0
for p in (3, 2):
s = g(a) + g(b) - g(c) - g(d)
q = p - 1
for i in range(abs(s)):
k += 1
if s > 0:
if g(a): a = a * q // p
else: b = b * q // p
else:
if g(c): c = c * q // p
else: d = d * q // p
if a * b != c * d: print(-1)
else: print(k, a, b, c, d)
``` | instruction | 0 | 76,243 | 24 | 152,486 |
Yes | output | 1 | 76,243 | 24 | 152,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
__author__ = 'zhan'
[a1, b1] = [int(i) for i in input().split()]
[a2, b2] = [int(i) for i in input().split()]
q = [[[a1, b1, 0]], [[a2, b2, 0]]]
total = [0, 0]
tested = [[], []]
def equal(t, q):
lo = 0
hi = len(q)
while True:
if lo >= hi:
return False
m = (lo + hi) // 2
p = q[m]
temp = p[0] * p[1]
if t == temp:
return [p[0], p[1], p[2]]
if t < temp:
lo = m + 1
elif t > temp:
hi = m
def expand(i):
ans = equal(total[i], q[(i+1)%2])
if ans:
print(
str(ans[2] + q[i][0][2]) + "\n" + str(q[i][0][0]) + " " + str(q[i][0][1]) + "\n" + str(ans[0]) + " " + str(
ans[1]))
else:
if not (q[i][0][0] & 1):
tt = [q[i][0][0] // 2, q[i][0][1], q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if q[i][0][0] % 3 == 0:
tt = [q[i][0][0] // 3 * 2, q[i][0][1], q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if not (q[i][0][1] & 1):
tt = [q[i][0][0], q[i][0][1] // 2, q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
if q[i][0][1] % 3 == 0:
tt = [q[i][0][0], q[i][0][1] // 3 * 2, q[i][0][2] + 1]
if not tt[0] * tt[1] in tested[i]:
q[i].append(tt)
tested[i].append(tt[0] * tt[1])
q[i].pop(0)
q[i].sort(key=lambda x: x[0] * x[1], reverse=True)
while True:
if len(q[0]) > 0 and len(q[1]) > 0:
total[0] = q[0][0][0] * q[0][0][1]
total[1] = q[1][0][0] * q[1][0][1]
if total[0] > total[1]:
expand(0)
elif total[0] < total[1]:
expand(1)
else:
print(str(q[0][0][2] + q[1][0][2]) + "\n" + str(q[0][0][0]) + " " + str(q[0][0][1]) + "\n" + str(
q[1][0][0]) + " " + str(q[1][0][1]))
break
else:
print(-1)
break
``` | instruction | 0 | 76,244 | 24 | 152,488 |
Yes | output | 1 | 76,244 | 24 | 152,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
def fact(a, b) :
ans = 0
while a % b == 0 :
ans += 1
a //= b
return ans
def fact_remove(a, b) :
c = a*b
while c % 2 == 0 : c //= 2
while c % 3 == 0 : c //= 3
return c
a1,b1 = map(int, input().split(' '))
a2,b2 = map(int, input().split(' '))
if fact_remove(a1, b1) != fact_remove(a2, b2) :
print(-1)
else :
ans = [0, 0, 0, 0]
c1 = a1*b1
c2 = a2*b2
k1 = fact(c1, 3)
k2 = fact(c2, 3)
if k1 > k2 :
ans[1] = k1 - k2
c1 /= 3**ans[1]
c1 *= 2**ans[1]
elif k1 < k2 :
ans[3] = k2 - k1
c2 /= 3**ans[3]
c2 *= 2**ans[3]
k1 = fact(c1, 2)
k2 = fact(c2, 2)
if k1 > k2 :
ans[0] = k1 - k2
c1 /= 2**ans[0]
elif k1 < k2 :
ans[2] = k2 - k1
c2 /= 2**ans[2]
if c1 != c2 :
print(-1)
else :
print(sum(ans))
while a1%3 == 0 and ans[1] > 0 :
a1 //= 3
a1 *= 2
ans[1] -= 1
while a1%2 == 0 and ans[0] > 0 :
a1 //= 2
ans[0] -= 1
while b1%3 == 0 and ans[1] > 0 :
b1 //= 3
b1 *= 2
ans[1] -= 1
while b1%2 == 0 and ans[0] > 0 :
b1 //= 2
ans[0] -= 1
while a2%3 == 0 and ans[3] > 0 :
a2 //= 3
a2 *= 2
ans[3] -= 1
while a2%2 == 0 and ans[2] > 0 :
a2 //= 2
ans[2] -= 1
while b2%3 == 0 and ans[3] > 0 :
b2 //= 3
b2 *= 2
ans[3] -= 1
while b2%2 == 0 and ans[2] > 0 :
b2 //= 2
ans[2] -= 1
print(a1, b1)
print(a2, b2)
``` | instruction | 0 | 76,245 | 24 | 152,490 |
Yes | output | 1 | 76,245 | 24 | 152,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
import sys
a1,b1 = map(int, input().split())
a2,b2 = map(int, input().split())
a, b = a1 * b1, a2 * b2
cnta2, cntb2, cnta3, cntb3 = 0, 0, 0, 0
ans = 0
while a%2==0:
a //= 2
cnta2 += 1
while a%3==0:
a //= 3
cnta3 += 1
while b%2==0:
b //= 2
cntb2 += 1
while b%3==0:
b //= 3
cntb3 += 1
if a != b:
print(-1)
sys.exit()
dif = cnta3 - cntb3
if dif > 0:
for i in range(dif):
ans += 1
if a1 % 3 == 0:
a1 = a1 * 2 // 3
else:
b1 = b1 * 2 // 3
else:
for i in range(-dif):
ans += 1
if a2 % 3 == 0:
a2 = a2 * 2 // 3
else:
b2 = b2 * 2 // 3
a, b = a1 * b1, a2 * b2
cnta, cntb = 0, 0
while a % 2 == 0:
a = a // 2
cnta += 1
while b % 2 == 0:
b = b // 2
cntb += 1
dif = cnta - cntb
if dif > 0:
for i in range(dif):
ans += 1
if a1 % 2 == 0:
a1 = a1 // 2
else:
b1 = b1 // 2
else:
for i in range(-dif):
ans += 1
if a2 % 2 == 0:
a2 = a2 // 2
else:
b2 = b2 // 2
print(ans)
print(str(a1) + ' ' + str(b1))
print(str(a2) + ' ' + str(b2))
``` | instruction | 0 | 76,246 | 24 | 152,492 |
Yes | output | 1 | 76,246 | 24 | 152,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
f = lambda: map(int, input().split())
a, b = f()
c, d = f()
p = [[a, b, c, d, 0, 0]]
n, q = -1, []
while p:
a, b, c, d, s, k = p.pop()
u, v = a * b, c * d
if u == v and (k < n or n < 0):
n, q = k, [a, b, c, d]
continue
if u > v: a, b, c, d, s = c, d, a, b, 1 - s
if c % 2 == 0: p.append((a, b, c // 2, d, s, k + 1))
if c % 3 == 0: p.append((a, b, 2 * c // 3, d, s, k + 1))
if d % 2 == 0: p.append((a, b, c, d // 2, s, k + 1))
if d % 3 == 0: p.append((a, b, c, 2 * d // 3, s, k + 1))
for t in [n] + q: print(t)
``` | instruction | 0 | 76,247 | 24 | 152,494 |
No | output | 1 | 76,247 | 24 | 152,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
"""
NTC here
"""
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
profile = 0
pypy = 1
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
def factors(a):
fact = []
if a % 2 == 0:
ch = 0
while a % 2 == 0:
ch += 1
a //= 2
fact.append([2, ch])
i = 3
if a % i == 0:
ch = 0
while a % i == 0:
ch += 1
a //= i
fact.append([i, ch])
return fact
from collections import defaultdict
def combine(a, b):
f1 = factors(a)
f2 = factors(b)
dc = defaultdict(int)
for i, j in f1:
dc[i] += j
for i, j in f2:
dc[i] += j
return dc
def main():
a1, b1 = lin()
a2, b2 = lin()
c1, c2= combine(a1, b1), combine(a2, b2)
sol = -1
ans = [-1]
def find(i, j, k, l):
if i-j<=c1[2] and k-l<=c2[2] and j<=c1[3] and l<=c2[3]:
x1 = (a1*b1*pow(2, j))//(pow(3, j)*pow(2, i))
x2 = (a2*b2*pow(2, l))//(pow(3, l)*pow(2, k))
# if x1==x2:print(x1, x2, i, j, k, l)
return x1 == x2
return False
for i in range(60):
for j in range(30):
for k in range(60):
for l in range(30):
if find(i, j, k, l):
if sol == -1 or sol>(i+j+k+l):
ans = [i, j, k, l]
sol = i+j+k+l
# print(ans)
print(sol)
if sol!=-1:
i, j, k, l = ans
# 3
while b1%3==0 and j:
b1//=3
b1*=2
j-=1
while b2%3==0 and l:
b2//=3
b2*=2
l-=1
while a1%3==0 and j:
a1//=3
a1*=2
j-=1
while a2%3==0 and l:
a2//=3
a2*=2
l-=1
# 2
while b1%2==0 and i:
b1//=2
i-=1
while b2%2==0 and k:
b2//=2
k-=1
while a1%2==0 and i:
a1//=2
i-=1
while a2%2==0 and k:
a2//=2
k-=1
# print(ans)
print(a1, b1)
print(a2, b2)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
if pypy:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
if profile:
import cProfile
cProfile.run('main()')
else:
main()
``` | instruction | 0 | 76,248 | 24 | 152,496 |
No | output | 1 | 76,248 | 24 | 152,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
a, b = map(int, input().split())
c, d = map(int, input().split())
s1 = a*b
s2 = c*d
mults = []
dc = {}
for i in range(65):
for j in range(65):
mult = 2**i*3**j
if mult <= s1 and mult <= s2:
mults.append(mult)
dc[mult] = [i, j, 2**j]
ans = []
mmin = 1e9
for i in mults:
for j in mults:
#print(i, j, end='')
x = sum(dc[i][:2]) + sum(dc[j][:2])
if x < mmin and (s1*dc[i][2])%i==0 and (s2*dc[j][2])%j==0 \
and s1//i*dc[i][2]==s2//j*dc[j][2]:
ans = [dc[i], dc[j]]
mmin = x
#print(' SUCESS')
if len(ans) == 0:
print(-1)
exit()
while ans[0][1] > 0:
if a % 3 == 0:
a //= 3
a *= 2
elif b % 3 == 0:
b //= 3
b *= 2
else:
print(-1)
exit()
ans[0][1] -= 1
while ans[0][0] > 0:
if a % 2 == 0:
a //= 2
elif b % 2 == 0:
b //= 2
else:
print(-1)
exit()
ans[0][0] -= 1
while ans[1][1] > 0:
if c % 3 == 0:
c //= 3
c *= 2
elif d % 3 == 0:
d //= 3
d *= 2
else:
print(-1)
exit()
ans[1][1] -= 1
while ans[1][0] > 0:
if c % 2 == 0:
c //= 2
elif d % 2 == 0:
d //= 2
else:
print(-1)
exit()
ans[1][0] -= 1
print(mmin)
print(a, b)
print(c, d)
``` | instruction | 0 | 76,249 | 24 | 152,498 |
No | output | 1 | 76,249 | 24 | 152,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented rectangle. The first bar is a1 Γ b1 segments large and the second one is a2 Γ b2 segments large.
Polycarpus wants to give Paraskevi one of the bars at the lunch break and eat the other one himself. Besides, he wants to show that Polycarpus's mind and Paraskevi's beauty are equally matched, so the two bars must have the same number of squares.
To make the bars have the same number of squares, Polycarpus eats a little piece of chocolate each minute. Each minute he does the following:
* he either breaks one bar exactly in half (vertically or horizontally) and eats exactly a half of the bar,
* or he chips of exactly one third of a bar (vertically or horizontally) and eats exactly a third of the bar.
In the first case he is left with a half, of the bar and in the second case he is left with two thirds of the bar.
Both variants aren't always possible, and sometimes Polycarpus cannot chip off a half nor a third. For example, if the bar is 16 Γ 23, then Polycarpus can chip off a half, but not a third. If the bar is 20 Γ 18, then Polycarpus can chip off both a half and a third. If the bar is 5 Γ 7, then Polycarpus cannot chip off a half nor a third.
What is the minimum number of minutes Polycarpus needs to make two bars consist of the same number of squares? Find not only the required minimum number of minutes, but also the possible sizes of the bars after the process.
Input
The first line of the input contains integers a1, b1 (1 β€ a1, b1 β€ 109) β the initial sizes of the first chocolate bar. The second line of the input contains integers a2, b2 (1 β€ a2, b2 β€ 109) β the initial sizes of the second bar.
You can use the data of type int64 (in Pascal), long long (in Π‘++), long (in Java) to process large integers (exceeding 231 - 1).
Output
In the first line print m β the sought minimum number of minutes. In the second and third line print the possible sizes of the bars after they are leveled in m minutes. Print the sizes using the format identical to the input format. Print the sizes (the numbers in the printed pairs) in any order. The second line must correspond to the first bar and the third line must correspond to the second bar. If there are multiple solutions, print any of them.
If there is no solution, print a single line with integer -1.
Examples
Input
2 6
2 3
Output
1
1 6
2 3
Input
36 5
10 16
Output
3
16 5
5 16
Input
3 5
2 1
Output
-1
Submitted Solution:
```
a, b = map(int, input().split())
c, d = map(int, input().split())
s1 = a*b
s2 = c*d
mults = []
dc = {}
for i in range(65):
for j in range(65):
mult = 2**i*3**j
if mult <= s1 and mult <= s2:
mults.append(mult)
dc[mult] = [i, j, 2**j]
ans = []
mmin = 1e9
for i in mults:
for j in mults:
#print(i, j, end='')
x = sum(dc[i][:2]) + sum(dc[j][:2])
if x < mmin and (s1*dc[i][2])%i==0 and (s2*dc[j][2])%j==0 \
and (s1*dc[i][2])//i==(s2*dc[j][2])//j:
ans = [dc[i], dc[j]]
mmin = x
#print(' SUCESS')
if len(ans) == 0:
print(-1)
exit()
while ans[0][1] > 0:
if a % 3 == 0:
a //= 3
a *= 2
elif b % 3 == 0:
b //= 3
b *= 2
else:
print(-1)
exit()
ans[0][1] -= 1
while ans[0][0] > 0:
if a % 2 == 0:
a //= 2
elif b % 2 == 0:
b //= 2
else:
print(-1)
exit()
ans[0][0] -= 1
while ans[1][1] > 0:
if c % 3 == 0:
c //= 3
c *= 2
elif d % 3 == 0:
d //= 3
d *= 2
else:
print(-1)
exit()
ans[1][1] -= 1
while ans[1][0] > 0:
if c % 2 == 0:
c //= 2
elif d % 2 == 0:
d //= 2
else:
print(-1)
exit()
ans[1][0] -= 1
print(mmin)
print(a, b)
print(c, d)
``` | instruction | 0 | 76,250 | 24 | 152,500 |
No | output | 1 | 76,250 | 24 | 152,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Methodius received an email from his friend Polycarp. However, Polycarp's keyboard is broken, so pressing a key on it once may cause the corresponding symbol to appear more than once (if you press a key on a regular keyboard, it prints exactly one symbol).
For example, as a result of typing the word "hello", the following words could be printed: "hello", "hhhhello", "hheeeellllooo", but the following could not be printed: "hell", "helo", "hhllllooo".
Note, that when you press a key, the corresponding symbol must appear (possibly, more than once). The keyboard is broken in a random manner, it means that pressing the same key you can get the different number of letters in the result.
For each word in the letter, Methodius has guessed what word Polycarp actually wanted to write, but he is not sure about it, so he asks you to help him.
You are given a list of pairs of words. For each pair, determine if the second word could be printed by typing the first one on Polycarp's keyboard.
Input
The first line of the input contains one integer n (1 β€ n β€ 10^5) β the number of pairs to check. Further input contains n descriptions of pairs.
The first line of each description contains a single non-empty word s consisting of lowercase Latin letters. The second line of the description contains a single non-empty word t consisting of lowercase Latin letters. The lengths of both strings are not greater than 10^6.
It is guaranteed that the total length of all words s in the input is not greater than 10^6. Also, it is guaranteed that the total length of all words t in the input is not greater than 10^6.
Output
Output n lines. In the i-th line for the i-th pair of words s and t print YES if the word t could be printed by typing the word s. Otherwise, print NO.
Examples
Input
4
hello
hello
hello
helloo
hello
hlllloo
hello
helo
Output
YES
YES
NO
NO
Input
5
aa
bb
codeforces
codeforce
polycarp
poolycarpp
aaaa
aaaab
abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz
Output
NO
NO
YES
NO
NO
Submitted Solution:
```
from collections import Counter
import math
from collections import Counter
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#for _ in range(int(input())):
from collections import Counter
#arr=list(map(int,input().split()))
n = int(input())
for _ in range(n):
s=input()
t=input()
i=0
for j in range(len(t)):
if i<len(s) and t[j]==s[i]:
i+=1
elif i>0 and t[j]==s[i-1]:
pass
else:
i=0
break
print("YES" if i==len(s) else "NO")
``` | instruction | 0 | 76,792 | 24 | 153,584 |
Yes | output | 1 | 76,792 | 24 | 153,585 |
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