message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 145 | 9 | 290 |
Tags: math
Correct Solution:
```
n = int(input())
for i in range(1, n + 1):
a, b, c = sorted(map(int, input().split()))
if c - a - b > 0:
print(a + b)
elif c - a - b == 0:
print(c)
else:
print(c+(a-c+b)//2)
``` | output | 1 | 145 | 9 | 291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 146 | 9 | 292 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
ll = list(map(int, input().split()))
ll.sort(reverse=True)
count = ll.pop()
x = (ll[0] - ll[1] + count)//2
if (ll[0] - count > ll[1]):
ll[0] -= count
else:
ll[0] -= x; ll[1] -= (count - x)
count += min(ll)
print(count)
``` | output | 1 | 146 | 9 | 293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 147 | 9 | 294 |
Tags: math
Correct Solution:
```
# Paulo Pacitti
# RA 185447
t = int(input())
for i in range(t):
case = [int(e) for e in input().split()]
case.sort(reverse=True)
days = 0
diff = case[0] - case[1]
if diff >= case[2]:
days += case[2]
case[0] -= case[2]
days += min(case[0], case[1])
else:
days += diff
case[0] -= diff
case[2] -= diff
days += (case[2] // 2) + case[1]
print(days)
``` | output | 1 | 147 | 9 | 295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 148 | 9 | 296 |
Tags: math
Correct Solution:
```
import math
for _ in range(int(input())):
l=list(map(int,input().split()))
l.sort()
if l[0]+l[1]<=l[2]:
print(l[0]+l[1])
else:
t=abs(l[1]+l[0]-l[2])
print(min(l[1]+l[0],l[2])+t//2)
``` | output | 1 | 148 | 9 | 297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 149 | 9 | 298 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
arr = [int(i) for i in input().split()]
arr.sort()
mn = arr[0]
diff = min(arr[2] - arr[1], arr[0])
x = (arr[0] - diff)//2
arr[1] -= x
arr[2] -= (arr[0] - x)
ans = min(arr[1], arr[2]) + mn
print(ans)
``` | output | 1 | 149 | 9 | 299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 150 | 9 | 300 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a,b,c=sorted(map(int,input().split()),reverse=True)
if a<=b+c:
print((a+b+c)//2)
else:
print(b+c)
``` | output | 1 | 150 | 9 | 301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 151 | 9 | 302 |
Tags: math
Correct Solution:
```
import math
N = int(input())
while N > 0:
a = input().split()
a[0] = int(a[0])
a[1] = int(a[1])
a[2] = int(a[2])
if a[0] > a[1]:
a[0],a[1] = a[1],a[0]
if a[0] > a[2]:
a[0],a[2] = a[2],a[0]
if a[1] > a[2]:
a[1],a[2] = a[2],a[1]
N-=1
if a[0] + a[1] >= a[2]:
print(math.floor((a[0] + a[1] + a[2]) / 2))
elif a[2] > a[0] + a[1]:
print(a[0] + a[1])
# print(a)
``` | output | 1 | 151 | 9 | 303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten. | instruction | 0 | 152 | 9 | 304 |
Tags: math
Correct Solution:
```
# balance
t = int(input())
for case in range(t):
arr = list(map(int, input().split()))
arr.sort()
d = arr[2] - arr[1]
if d >= arr[0]:
print(arr[0] + arr[1])
else: # d < arr[0]
arr[0] -= d
# arr[2] = arr[1]
if arr[0] & 1:
print(d + arr[1] + arr[0] - (arr[0] >> 1) - 1)
else:
print(d + arr[1] + arr[0] - (arr[0] >> 1))
``` | output | 1 | 152 | 9 | 305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
for i in range(int(input())):
l=sorted(list(map(int,input().split())))
if l[0]+l[1]<=l[2]:
print(l[0]+l[1])
else:
print(sum(l)//2)
``` | instruction | 0 | 153 | 9 | 306 |
Yes | output | 1 | 153 | 9 | 307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
n = int(input())
for i in range(n):
a,b,c = map(int,input().split())
a,b,c = sorted([a,b,c])
res = 0
if c >= b - a:
res += b - a
c -= b - a
b = a
mx = min(a, c // 2)
res += mx * 2
c -= mx * 2
a -= mx
b -= mx
res += a
print(res)#, a, b, c)
``` | instruction | 0 | 154 | 9 | 308 |
Yes | output | 1 | 154 | 9 | 309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
for _ in range(int(input())):
l = list(map(int,input().split()))
l.sort()
print(min(sum(l)//2,l[0]+l[1]))
``` | instruction | 0 | 155 | 9 | 310 |
Yes | output | 1 | 155 | 9 | 311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
T = int(input())
def solve(r, g, b):
r, g, b = list(sorted([r, g, b]))
#print(r, g, b)
eaten = 0
eaten += g-r
r, g, b = r, g-eaten, b-eaten
#print(r, g, b, "eaten = ", eaten)
k = min(b - g, g, b//2)
eaten += 2*k
r, g, b = r-k, g-k, b-2*k # b - 2*(b - g) =
#print(r, g, b, "eaten = ", eaten)
if r == 0: return eaten
else:
assert r == b
return eaten + (r+g+b) // 2
#print("----")
def sim(r, g, b):
r, g, b = list(sorted([r, g, b]))
eaten = 0
while g != 0:
g, b = g-1, b-1
eaten += 1
r, g, b = list(sorted([r, g, b]))
return eaten
#for r in range(100):
# for g in range(100):
# for b in range(100):
# assert sim(r, g, b) == solve(r, g, b)
#print("sim ok")
for _ in range(T):
r, g, b = [int(x) for x in input().split()]
print(solve(r, g, b))
``` | instruction | 0 | 156 | 9 | 312 |
Yes | output | 1 | 156 | 9 | 313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
t = int(input())
for i in range (t):
a=sorted(list(map(int,input().split())))
if a[0]==a[1] and a[1]==a[2] and a[0]==a[2]:
print(int(a[0]))
continue
elif a[1]==a[2]:
print(a[0]+min(a[2]-int(a[0]/2),a[1]-(a[0]%2)))
#print(int(a[0]+((2*a[1]-a[0])/2)))
continue
else:
print(a[1]+min(a[0],a[2]-a[1]))
continue
``` | instruction | 0 | 157 | 9 | 314 |
No | output | 1 | 157 | 9 | 315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
t=int(input())
for i in range(0,t):
r=list(map(int,input().split()))
if(((r[0]==r[1]) and (r[1]==r[2]))or((r.count(1)==0)and(sum(r)%2==0))):
s=(r[0]+r[1]+r[2])//2
print(s)
continue
else:
if(r.count(max(r))==2):
print(sum(r)//2)
continue
else:
k=r[r.index(max(r))]
r[r.index(max(r))]=0
print(min(sum(r),k))
continue
``` | instruction | 0 | 158 | 9 | 316 |
No | output | 1 | 158 | 9 | 317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
t = int(input())
for i in range(t):
ans = 0
a = list(map(int, input().split()))
if a.count(max(a)) == 1:
mi = min(a)
a.pop(a.index(mi))
ma = max(a)
a[a.index(ma)] -= mi
ans += mi
mi = min(a)
ans += mi
elif a.count(max(a)) == 3:
ans += sum(a) // 2
else:
mi = min(a)
a.pop(a.index(mi))
mi1 = mi // 2
a[0] -= mi1
a[1] -= mi1
a[0] -= (mi % 2)
ans += mi + min(a)
print(ans)
``` | instruction | 0 | 159 | 9 | 318 |
No | output | 1 | 159 | 9 | 319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have three piles of candies: red, green and blue candies:
* the first pile contains only red candies and there are r candies in it,
* the second pile contains only green candies and there are g candies in it,
* the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Then t test cases follow.
Each test case is given as a separate line of the input. It contains three integers r, g and b (1 ≤ r, g, b ≤ 10^8) — the number of red, green and blue candies, respectively.
Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.
Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
Submitted Solution:
```
for _ in range(int(input())):
a = list(map(int,input().split()))
a.sort()
ans = 0
ans+= a[1]-1
a[2] -= a[1]-1
a[1]=1
a.sort()
ans+= a[1]-1
a[2] -= a[1]-1
a[1]=1
if a[2]>=2:
ans+=2
else:
ans+=1
print(ans)
``` | instruction | 0 | 160 | 9 | 320 |
No | output | 1 | 160 | 9 | 321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 343 | 9 | 686 |
Tags: greedy, implementation
Correct Solution:
```
MOD = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
n,v=f()
l=[0]*3003
for _ in range(n):
a,b=f()
l[a]+=b
fromprevDay=0
currDay=0
temp=0
ans=0
for i in range(1,3002):
currDay=l[i]
if fromprevDay+currDay<=v:
ans+=fromprevDay+currDay
fromprevDay=0
elif fromprevDay>=v:
ans+=v
fromprevDay=currDay
else:
ans+=v
temp=v-fromprevDay
fromprevDay=currDay-temp
print(ans)
``` | output | 1 | 343 | 9 | 687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 344 | 9 | 688 |
Tags: greedy, implementation
Correct Solution:
```
n,v=map(int, input().split())
b=[0]*3002
for _ in range(n):
a = list(map(int,input().split()))
b[a[0]]+=a[1]
s = 0
for i in range(1,3002):
r=v
t=min(r,b[i-1])
s+=t
r-=t
b[i-1]-=t
t=min(r, b[i])
s+=t
r-=t
b[i]-=t
print(s)
``` | output | 1 | 344 | 9 | 689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 345 | 9 | 690 |
Tags: greedy, implementation
Correct Solution:
```
import sys
import math
n, v = [int(x) for x in (sys.stdin.readline()).split()]
k = [0] * 3002
for i in range(n):
a, b = [int(x) for x in (sys.stdin.readline()).split()]
k[a] += b
i = 1
res = 0
while(i <= 3001):
val = 0
if(k[i - 1] < v):
val = k[i - 1]
res += val
d = v - val
if(k[i] < d):
res += k[i]
k[i] = 0
else:
res += d
k[i] = k[i] - d
else:
res += v
i += 1
print(res)
``` | output | 1 | 345 | 9 | 691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 346 | 9 | 692 |
Tags: greedy, implementation
Correct Solution:
```
n, v = map(int, input().split())
d, s = {}, 0
for i in range(n):
a, b = map(int, input().split())
d[a] = d.get(a, 0) + b
s += b
r = v
for i in sorted(d.keys()):
d[i] -= min(d[i], r)
r, b = v, min(d[i], v)
d[i] -= b
if i+1 in d:
r -= b
print(s - sum(d.values()))
``` | output | 1 | 346 | 9 | 693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 347 | 9 | 694 |
Tags: greedy, implementation
Correct Solution:
```
from collections import defaultdict
dd = defaultdict(int)
n, v = map(int, input().split())
def foo(day, toadd):
z = 0
if(newDD[day]<toadd):
z+=newDD[day]
newDD[day] = 0
else:
z+=toadd
newDD[day]-=toadd
return z
for _ in range(n):
temp1, temp2 = map(int, input().split())
dd[temp1]+=temp2
newDD = dict(sorted(dd.items()))
# print(newDD)
ans = 0
mxday = max(newDD.keys())
for key in range(1, mxday+2):
V = v
if((key-1) in newDD):
if(newDD[key-1]>0):
temp=foo(key-1, V)
V-=temp
ans+=temp
if(key in newDD):
if(V>0):
ans+=foo(key, V)
print(ans)
``` | output | 1 | 347 | 9 | 695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 348 | 9 | 696 |
Tags: greedy, implementation
Correct Solution:
```
n, v = map(int, input().split(' '))
trees = dict()
count = 0
b_last = 0
for i in range(n):
a, b = map(int, input().split(' '))
if trees.get(a):
trees[a] += b
else:
trees[a] = b
m = max(trees.keys())
for i in range(1, m+2):
if trees.get(i):
k = min(v, b_last)
count += k
k1 = min(v - k, trees[i])
count += k1
b_last = trees[i] - k1
else:
count += min(v, b_last)
b_last = 0
print(count)
``` | output | 1 | 348 | 9 | 697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 349 | 9 | 698 |
Tags: greedy, implementation
Correct Solution:
```
def main():
n,m = map(int,input().split())
a = []
b = []
for _ in range(n):
a1,b1 = map(int,input().split())
a.append(a1)
b.append(b1)
prev = 0
tv = 0
ans = 0
for i in range(1,3002):
curr = 0
for j in range(n):
if a[j]==i:
curr+=b[j]
if curr+prev<=m:
ans += prev+curr
prev = 0
else:
ans += m
tv = m - prev
if tv<0:
tv = 0
prev = curr-tv
print(ans)
return
main()
``` | output | 1 | 349 | 9 | 699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither. | instruction | 0 | 350 | 9 | 700 |
Tags: greedy, implementation
Correct Solution:
```
n,v = [int(i) for i in input().split()]
f = 0
l = [0]*3002
for i in range(n):
a,b = [int(i) for i in input().split()]
l[a] += b
for i in range(1,3002):
pick = min(v,l[i-1])
l[i-1] -= pick
pick2 = min(v-pick,l[i])
l[i] -= pick2
f += pick + pick2
print(f)
``` | output | 1 | 350 | 9 | 701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
def main() :
R = lambda : map(int, input().split())
n, v = R()
mlist = [0] * 3003
for i in range(n) :
a, b = R()
mlist[a] += b
mlist2 = mlist[:]
ans = 0
for i in range(1, 3002) :
tmp = min(v, mlist[i])
ans += tmp
mlist[i + 1] += min(mlist[i] - tmp, mlist2[i])
print(ans)
if __name__ == "__main__" :
main()
``` | instruction | 0 | 351 | 9 | 702 |
Yes | output | 1 | 351 | 9 | 703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
n,v = map(int, input().split())
arr = [0]*3002
for _ in range(n):
a,b = map(int, input().split())
arr[a] += b
x = v
ans = 0
prev = 0
for i in range(1,3002):
x = max(x-prev,0)
ans += (v-x)
if arr[i]<=x:
ans += arr[i]
prev = 0
else:
ans += x
prev = arr[i]-x
x = v
print(ans)
``` | instruction | 0 | 352 | 9 | 704 |
Yes | output | 1 | 352 | 9 | 705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def main():
try:
n, v = invr()
a = [0]*(3002)
for i in range(n):
x, y = invr()
a[x] += y
ans = 0
prev = 0
for i in range(1, 3002):
curr = a[i]
if curr + prev <= v:
ans += curr + prev
curr = 0
else:
ans += v
if prev < v:
curr -= v - prev
prev = curr
print(ans)
except Exception as e:
print(e)
# 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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 353 | 9 | 706 |
Yes | output | 1 | 353 | 9 | 707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
n, v = map(int, input().split())
T = 3010
a = [[] for i in range(T)]
for i in range(n):
x, y = map(int, input().split())
a[x].append([y, 1])
ans = 0
for i in range(T):
a[i] = a[i][::-1]
j = 0
cur = v
while j < len(a[i]):
if a[i][j][0] > cur:
ans += cur
if a[i][j][1] == 0:
j += 1
else:
a[i][j][0] -= cur
break
else:
ans += a[i][j][0]
cur -= a[i][j][0]
j += 1
for t in range(j, len(a[i])):
if a[i][t][1] != 0:
a[i + 1].append([a[i][t][0], 0])
print(ans)
``` | instruction | 0 | 354 | 9 | 708 |
Yes | output | 1 | 354 | 9 | 709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
import sys
import math
n, v = [int(x) for x in (sys.stdin.readline()).split()]
k = [0] * (n + 2)
for i in range(n):
a, b = [int(x) for x in (sys.stdin.readline()).split()]
k[a] = b
i = 1
res = 0
while(i <= n + 1):
val = 0
if(k[i - 1] < v):
val = k[i - 1]
res += val
d = v - val
if(k[i] < d):
res += k[i]
k[i] = 0
else:
res += d
k[i] = k[i] - d
else:
res += v
i += 1
print(res)
``` | instruction | 0 | 355 | 9 | 710 |
No | output | 1 | 355 | 9 | 711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
n,v = map(int,input().split())
lis=[0]*(4000)
for i in range(n):
a,b = map(int,input().split())
lis[a]+=b
prev=0
ans=0
for i in range(1,3000):
cap=v
# print(prev,lis[i],v)
cap=v-min(prev,v)
ans+=min(prev,v)
zz=min(lis[i],cap)
ans+=zz
prev=lis[i]-zz
# print(ans)
print(ans)
``` | instruction | 0 | 356 | 9 | 712 |
No | output | 1 | 356 | 9 | 713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
n, v = map(int, input().split())
l = [0]*(3010)
ans = 0
for i in range(n):
a, b = map(int, input().split())
l[a] = b
day = 1
for i in range(3004):
temp = min(v, l[day-1])
ans += temp
l[day-1] -= temp
rem = v - temp
temp2 = min(rem, l[day])
ans+=temp2
l[day]-=temp2
day+=1
print(ans)
``` | instruction | 0 | 357 | 9 | 714 |
No | output | 1 | 357 | 9 | 715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
Input
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60
Note
In the first sample, in order to obtain the optimal answer, you should act as follows.
* On the first day collect 3 fruits from the 1-st tree.
* On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
* On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
Submitted Solution:
```
def read_integer_pair():
line = input()
pair = line.split(" ")
assert len(pair) == 2
return int(pair[0]), int(pair[1])
if __name__ == '__main__':
n, v = read_integer_pair()
trees = {}
for i in range(n):
a, b = read_integer_pair()
if a in trees:
trees[a] += b
else:
trees[a] = b
trees = list(trees.items())
trees.sort(key=lambda x: x[0])
prev_a = 0
prev_b = 0
apples = 0
for a, b in trees:
dv = min(v, prev_b)
apples += dv
if a - prev_a != 1:
dv = 0
if dv < v:
dv = min(v - dv, b)
apples += dv
prev_a = a
prev_b = b - dv
dv = min(v, prev_b)
apples += dv
print(apples)
``` | instruction | 0 | 358 | 9 | 716 |
No | output | 1 | 358 | 9 | 717 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 788 | 9 | 1,576 |
"Correct Solution:
```
from itertools import accumulate
n, t = [int(x) for x in input().split()]
T = [0]*(t+1)
for _ in range(n):
l, r = [int(x) for x in input().split()]
T[l] += 1
T[r] -= 1
print(max(accumulate(T)))
``` | output | 1 | 788 | 9 | 1,577 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 790 | 9 | 1,580 |
"Correct Solution:
```
from itertools import accumulate
n, t = map(int, input().split())
a = [0] * (t + 1)
for l, r in (map(int, input().split()) for _ in range(n)):
a[l] += 1
a[r] -= 1
print(max(accumulate(a)))
``` | output | 1 | 790 | 9 | 1,581 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 793 | 9 | 1,586 |
"Correct Solution:
```
N, T = map(int, input().split())
D = [0]*(T+1)
for i in range(N):
l, r = map(int, input().split())
D[l] += 1; D[r] -= 1
for i in range(T):
D[i+1] += D[i]
print(max(D))
``` | output | 1 | 793 | 9 | 1,587 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.
Constraints
* $ 1 \leq N \leq 10^5 $
* $ 1 \leq T \leq 10^5 $
* $ 0 \leq l_i < r_i \leq T $
Input
The input is given in the following format.
$N$ $T$
$l_1$ $r_1$
$l_2$ $r_2$
:
$l_N$ $r_N$
Output
Print the maximum number of persons in a line.
Examples
Input
6 10
0 2
1 3
2 6
3 8
4 10
5 10
Output
4
Input
2 2
0 1
1 2
Output
1 | instruction | 0 | 794 | 9 | 1,588 |
"Correct Solution:
```
import heapq
from collections import deque
from enum import Enum
import sys
import math
from _heapq import heappush, heappop
#------------------------------------------#
BIG_NUM = 2000000000
HUGE_NUM = 9999999999999999
MOD = 1000000007
EPS = 0.000000001
#------------------------------------------#
N,T = map(int,input().split())
table = [0]*(T+1)
for loop in range(N):
left,right = map(int,input().split())
table[left] += 1
table[right] -= 1
ans = table[0]
for i in range(1,T+1):
table[i] += table[i-1]
ans = max(ans,table[i])
print("%d"%(ans))
``` | output | 1 | 794 | 9 | 1,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,301 | 9 | 2,602 |
Tags: implementation, math
Correct Solution:
```
n,h,k = map(int,input().split())
arr = [int(x) for x in input().split()]
ph = 0
second = 0
for i in range(n):
ph += arr[i]
if(i==n-1):
break
d = h-ph
x = arr[i+1]-d
if(x<=0):
continue
plus = 0
if(x%k==0):
plus = int(x/k)
else:
plus = int(x/k)+1
minus = k*plus
ph -= minus
if(ph<0):
ph = 0
second += plus
plus = 0
if(ph%k==0):
plus = int(ph/k)
else:
plus = int(ph/k)+1
second += plus
print(second)
``` | output | 1 | 1,301 | 9 | 2,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,302 | 9 | 2,604 |
Tags: implementation, math
Correct Solution:
```
d=input().split()
d=[int(x) for x in d]
n,h,k=d[0],d[1],d[2]
d=input().split()
d=[int(x) for x in d]
S=0
R=0
for i in d:
if R+i<=h:
S+=i//k
R+=i%k
else:
while R+i>h:
if R%k==0:
S+=R//k
R=0
elif R<k:
S+=1
R=0
else:
S+=R//k
R=R%k
S+=i//k
R+=i%k
if R%k==0:
S+=R//k
else:
S+=R//k
S+=1
print(S)
``` | output | 1 | 1,302 | 9 | 2,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,303 | 9 | 2,606 |
Tags: implementation, math
Correct Solution:
```
n,h,k=map(int,input().split())
ip=list(map(int,input().split()))
time=0
count=0
height=0
while count<n:
if height + ip[count]<= h:
height+=ip[count]
count+=1
else:
if height%k==0:
time+=height//k
height=0
elif height>=k:
time+=height//k
height=height%k
else:
height-=min(k,height)
time+=1
#print(height,time,count)
if height==0:
s=0
elif height%k==0:
s=height//k
else:
s=(height//k)+1
print(time+s)
``` | output | 1 | 1,303 | 9 | 2,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,304 | 9 | 2,608 |
Tags: implementation, math
Correct Solution:
```
from math import ceil
n, h, k = map(int, input().strip().split())
potatoes = map(int, input().strip().split())
sec = 0
in_progress = 0
for p in potatoes:
while p + in_progress > h:
if in_progress < k:
sec += 1
in_progress = 0
else:
elapsed, in_progress = divmod(in_progress, k)
sec += elapsed
in_progress += p
sec += ceil(in_progress/k)
print(sec)
``` | output | 1 | 1,304 | 9 | 2,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,305 | 9 | 2,610 |
Tags: implementation, math
Correct Solution:
```
n, mxH, k = map(int, input().split())
arr = list(map(int, input().split()))
currT, currH, res = 0, 0, 0
i = 0
while i < len(arr):
while (i < len(arr)) and (currH + arr[i] <= mxH):
currH += arr[i]
i += 1
if i < len(arr):
currT = (currH + arr[i] - mxH + k - 1) // k
else:
currT = (currH + k - 1) // k
currH = max(currH - currT * k, 0)
res += currT
print(res)
``` | output | 1 | 1,305 | 9 | 2,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,306 | 9 | 2,612 |
Tags: implementation, math
Correct Solution:
```
n, h, k = map(int,input().split())
t, x = 0, 0
w=list(map(int,input().split()))
for z in w:
t += x // k; x %= k
if x + z > h: t, x = t + 1, 0
x += z
t+=(x+k-1)//k
print(t)
``` | output | 1 | 1,306 | 9 | 2,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,307 | 9 | 2,614 |
Tags: implementation, math
Correct Solution:
```
import math
if __name__ == '__main__':
n, h, k = map(int, input().split())
heights = list(map(int, input().split()))
time = 0
r = 0
i = 0
while i < n:
while i < n and h - r >= heights[i]:
r += heights[i]
i += 1
if i < n:
a = math.ceil((heights[i] - (h - r)) / k)
r -= a * k
time += a
if r < 0:
r = 0
time += math.ceil(r / k)
print(time)
``` | output | 1 | 1,307 | 9 | 2,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds. | instruction | 0 | 1,308 | 9 | 2,616 |
Tags: implementation, math
Correct Solution:
```
######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
######### # # # # ##### # ##### # ## # ## # #
"""
PPPPPPP RRRRRRR OOOO VV VV EEEEEEEEEE
PPPPPPPP RRRRRRRR OOOOOO VV VV EE
PPPPPPPPP RRRRRRRRR OOOOOOOO VV VV EE
PPPPPPPP RRRRRRRR OOOOOOOO VV VV EEEEEE
PPPPPPP RRRRRRR OOOOOOOO VV VV EEEEEEE
PP RRRR OOOOOOOO VV VV EEEEEE
PP RR RR OOOOOOOO VV VV EE
PP RR RR OOOOOO VV VV EE
PP RR RR OOOO VVVV EEEEEEEEEE
"""
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
from bisect import bisect_left
def solve():
n, h, k = map(int, input().split())
a = list(map(int, input().split()))
c = 0
summ = 0
waste = 0
for x in a:
summ += c // k
c %= k
if c + x > h:
summ, c = summ + 1, 0
c += x
print(summ + math.ceil(c / k))
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | output | 1 | 1,308 | 9 | 2,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
n, h, j = map(int, input().split())
pots = [int(x) for x in input().split()]
count = 0
height = 0
pot = 0
while pot < n:
if height + pots[pot] <= h:
height += pots[pot]
count += height // j
height = height % j
pot += 1
elif height <= j:
height = 0
count += 1
else:
count += height // j
height = height % j
if height > 0:
print(count +1)
else:
print(count)
``` | instruction | 0 | 1,309 | 9 | 2,618 |
Yes | output | 1 | 1,309 | 9 | 2,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
import sys
n, h, k = map(int, sys.stdin.readline().split())
an = list(map(int, sys.stdin.readline().split()))
ans = 0
cur = 0
height = 0
while cur < n:
if height + an[cur] <= h:
height += an[cur]
else:
ans += 1
height = an[cur]
ans += height // k
height %= k
cur += 1
if cur == n and height != 0:
ans += 1
print(ans)
``` | instruction | 0 | 1,310 | 9 | 2,620 |
Yes | output | 1 | 1,310 | 9 | 2,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
n,h,k=map(int,input().split())
xs=list(map(int,input().split()))+[h+1]
t=0
x=0
i=0
while i+1<len(xs) or x>0:
while x+xs[i]<=h:
x+=xs[i]
i+=1
d=max(1,min(x,x-h+xs[i]+k-1)//k)
x=max(0,x-d*k)
t+=d
print(t)
``` | instruction | 0 | 1,311 | 9 | 2,622 |
Yes | output | 1 | 1,311 | 9 | 2,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
# 677B
# θ(n) time
# θ(n) space
__author__ = 'artyom'
# SOLUTION
def main():
n, h, k = read(3)
a = read(3)
c = i = d = 0
while i < n:
while i < n and d + a[i] <= h:
d += a[i]
i += 1
x = a[i] - h + d if i < n else d
t = (x - 1) // k + 1
d = max(0, d - t * k)
c += t
return c
# HELPERS
def read(mode=1, size=None):
# 0: String
# 1: Integer
# 2: List of strings
# 3: List of integers
# 4: Matrix of integers
if mode == 0: return input().strip()
if mode == 1: return int(input().strip())
if mode == 2: return input().strip().split()
if mode == 3: return list(map(int, input().strip().split()))
a = []
for _ in range(size):
a.append(read(3))
return a
def write(s="\n"):
if s is None: s = ''
if isinstance(s, tuple) or isinstance(s, list): s = ' '.join(map(str, s))
s = str(s)
print(s, end="\n")
write(main())
``` | instruction | 0 | 1,312 | 9 | 2,624 |
Yes | output | 1 | 1,312 | 9 | 2,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
import sys
n, h, k = map(int,sys.stdin.readline().split())
papa = input()
#n, h, k = map(int,"5 6 3".split())
#papa = "5 4 3 2 1"
#papa = "5 5 5 5 5"
#papa = "1 2 1 1 1"
papa = papa.split(' ')
papa = [int(i) for i in papa]
papa.append(2 * k)
print (n, h, k, papa)
sec = 0
food = 0
while len(papa) > 1:# and food > 0:
#print (n, h, k, food, sec, papa)
while food + papa[0] <= h:
food += papa[0]
del papa[0]
food -= k
if food < 0:
food = 0
sec += 1
#print (food, sec)
while food > 0:
food -= k
if food < 0:
food = 0
sec += 1
print (sec)
``` | instruction | 0 | 1,313 | 9 | 2,626 |
No | output | 1 | 1,313 | 9 | 2,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
2. Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
Input
The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
Output
Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examples
Input
5 6 3
5 4 3 2 1
Output
5
Input
5 6 3
5 5 5 5 5
Output
10
Input
5 6 3
1 2 1 1 1
Output
2
Note
Consider the first sample.
1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
Submitted Solution:
```
n, h, k = list(map(int, input().split()))
a = list(reversed(sorted(list(map(int, input().split())))))
out = 0
cur = 0
while len(a) > 0:
tmp = cur
i = 0
while i < len(a):
if tmp + a[i] <= h:
tmp+=a[i]
del(a[i])
i-=1
i+=1
cur = tmp
if cur < k:
cur = 0
else:
cur-=k
out+=1
print(out+1)
``` | instruction | 0 | 1,314 | 9 | 2,628 |
No | output | 1 | 1,314 | 9 | 2,629 |
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