message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A pair of positive integers (a,b) is called special if ⌊ a/b ⌋ = a mod b. Here, ⌊ a/b ⌋ is the result of the integer division between a and b, while a mod b is its remainder.
You are given two integers x and y. Find the number of special pairs (a,b) such that 1≤ a ≤ x and 1 ≤ b ≤ y.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The only line of the description of each test case contains two integers x, y (1 ≤ x,y ≤ 10^9).
Output
For each test case print the answer on a single line.
Example
Input
9
3 4
2 100
4 3
50 3
12 4
69 420
12345 6789
123456 789
12345678 9
Output
1
0
2
3
5
141
53384
160909
36
Note
In the first test case, the only special pair is (3, 2).
In the second test case, there are no special pairs.
In the third test case, there are two special pairs: (3, 2) and (4, 3).
Submitted Solution:
```
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
# ---------------------------- template ends here ----------------------------
def solve_(a,b):
# your solution here
# a > b
# a < b**2
res = 0
for y in range(2, min(int(10**5) + 10, b+1)):
cnt = min(y-1, a // (y + 1))
# log(y,cnt)
res += cnt
# check_res = 0
# for i in range(1,a+1):
# for j in range(1,b+1):
# if i//j == i%j:
# check_res += 1
# log(res, check_res)
# assert res == check_res
return res
# for p in range(1,1000):
# for q in range(1,1000):
# solve(q,p)
# solve(10**9, 10**9)
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
a,b = list(map(int,input().split()))
# lst = list(map(int,input().split()))
# read multiple rows
# mrr = read_matrix(k) # and return as a list of list of int
# arr = read_strings(k) # and return as a list of str
res = solve(a,b) # include input here
# print result
# Google and Facebook - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Other platforms - no case number required
print(res)
# print(len(res))
# print(*res) # print a list with elements
# for r in res: # print each list in a different line
# print(res)
# print(*res)
``` | instruction | 0 | 46,510 | 22 | 93,020 |
No | output | 1 | 46,510 | 22 | 93,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A pair of positive integers (a,b) is called special if ⌊ a/b ⌋ = a mod b. Here, ⌊ a/b ⌋ is the result of the integer division between a and b, while a mod b is its remainder.
You are given two integers x and y. Find the number of special pairs (a,b) such that 1≤ a ≤ x and 1 ≤ b ≤ y.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The only line of the description of each test case contains two integers x, y (1 ≤ x,y ≤ 10^9).
Output
For each test case print the answer on a single line.
Example
Input
9
3 4
2 100
4 3
50 3
12 4
69 420
12345 6789
123456 789
12345678 9
Output
1
0
2
3
5
141
53384
160909
36
Note
In the first test case, the only special pair is (3, 2).
In the second test case, there are no special pairs.
In the third test case, there are two special pairs: (3, 2) and (4, 3).
Submitted Solution:
```
from math import ceil, floor
def func(x, y):
if x <= 1000 or y <= 1000:
ans = 0
for i in range(2, y + 1):
ans += min(x // (i + 1), i - 1)
if i > x:
return ans
return ans
ans = 0
# cnt = {}
# xx = 0
for i in range(2, y + 1):
if i - 1 >= x // (i + 1):
# tmp = x // (i + 1)
# ans += tmp
# xx += tmp
# if tmp in cnt: cnt[tmp] += 1
# if tmp not in cnt: cnt[tmp] = 1
break
else:
tmp = i - 1
ans += tmp
# print(i)
# for i in cnt:
# print(i, cnt[i])
# print(xx)
# return ans
# print(ans)
if i - 1 < x // (i + 1):
return ans
MAX = x // (i + 1)
MIN = x // (y + 1)
# print(MAX, MIN)
xx = 0
for i in range(MAX, MIN, -1):
ans += i * (floor(x / i) - ceil(x / (i + 1)) + 1)
if x % (i + 1) == 0:
ans -= i
# print(i, (floor(x / i) - ceil(x / (i + 1)) ) )
# else:
# print(i, (floor(x / i) - ceil(x / (i + 1)) + 1))
j = y
ans += MIN * (y - ceil(x / (MIN + 1)) + 1)
if x % (MIN + 1): ans -= MIN
return ans
n = int(input())
for i in range(n):
x, y = input().split()
x = int(x)
y = int(y)
print(func(x, y))
``` | instruction | 0 | 46,511 | 22 | 93,022 |
No | output | 1 | 46,511 | 22 | 93,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,721 | 22 | 93,442 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n, m = map(int, input().split(" "))
counter = 0
for i in range(1, m+1):
no = i % 5
counter += (n+(i % 5))//5
print(counter)
``` | output | 1 | 46,721 | 22 | 93,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,722 | 22 | 93,444 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
x,y = map(int,input().split())
arx = []
ary = []
xx = x//5
arx = [xx,xx,xx,xx,xx]
for i in range(x%5):
arx[i] += 1
yy = y//5
ary = [yy,yy,yy,yy,yy]
for i in range(y%5):
ary[i] += 1
sum = 0
sum += arx[0]*ary[3]
sum += arx[1]*ary[2]
sum += arx[2]*ary[1]
sum += arx[3]*ary[0]
sum += arx[4]*ary[4]
print(sum)
``` | output | 1 | 46,722 | 22 | 93,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,723 | 22 | 93,446 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n, m = map(int, input().split())
cnt = 0
N = 5 * int(1e5)
for k in range(N):
c = k * 5
Max = min(n, c - 1)
Min = max(1, c - m)
if Max >= c - m and Min <= c - 1:
cnt += Max - Min + 1
print(cnt)
``` | output | 1 | 46,723 | 22 | 93,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,724 | 22 | 93,448 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n,m = list(map(int,input().split()))
cm = {}
cn = {}
x = n//5
y = m//5
for i in range(5):
cm[i]=y
cn[i]=x
x = n%5
y = m%5
for i in range(1,x+1):
cn[i]+=1
for i in range(1,y+1):
cm[i]+=1
res = cm[0]*cn[0]
for i in range(1,5):
res+=cm[i]*cn[5-i]
print(res)
``` | output | 1 | 46,724 | 22 | 93,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,725 | 22 | 93,450 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n,m = list(map(int,input().split()))
S=0
for i in range(1,n+1):
S+=((m-(5*(i//5)+5)+i)//5)+1
print(S)
``` | output | 1 | 46,725 | 22 | 93,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,726 | 22 | 93,452 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n,m = [int(x) for x in input().split()]
freq1 = [0 for x in range(5)]
freq2 = [0 for x in range(5)]
for i in range(1,n+1):
freq1[i%5]+=1
for i in range(1,m+1):
freq2[i%5]+=1
c=0
for i in range(5):
c+= freq1[i%5]*freq2[abs(5-i)%5]
print(c)
``` | output | 1 | 46,726 | 22 | 93,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,727 | 22 | 93,454 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n, m = map(int, input().split())
nt = n // 5
mt = m // 5
nf = [nt] * 5
mf = [mt] * 5
for x in range((n % 5) + 1):
nf[x] += 1
for x in range((m % 5) + 1):
mf[x] += 1
nf[0] -= 1
mf[0] -= 1
ans = 0
for x in range(5):
y = (5 - x) % 5
ans += nf[x] * mf[y]
print(ans)
``` | output | 1 | 46,727 | 22 | 93,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | instruction | 0 | 46,728 | 22 | 93,456 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
n, m = map(int, input().split(" "))
res = 0
print(sum((((i+1)%5+m)//5)*(n//5+1) for i in range(5) if n%5>=(i+1))+
sum((((i+1)%5+m)//5)*(n//5) for i in range(5) if n%5<(i+1)))
``` | output | 1 | 46,728 | 22 | 93,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
n, m = map(int, input().split(" "))
counter = 0
for i in range(1, m+1):
counter += (n+(i % 5))//5
print(counter)
``` | instruction | 0 | 46,729 | 22 | 93,458 |
Yes | output | 1 | 46,729 | 22 | 93,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
#n = int(input())
n, m = map(int, input().split())
#s = input()
#c = list(map(int, input().split()))
l = (n // 5) * (m // 5)
for i in range(1, 5):
l += ((n - i) // 5 + 1) * ((m - 5 + i) // 5 + 1)
print(l)
``` | instruction | 0 | 46,730 | 22 | 93,460 |
Yes | output | 1 | 46,730 | 22 | 93,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
import sys
def main():
x = sys.stdin.readline().split()
n, m = int(x[0]), int(x[1])
k = int(n/5)
rest = n - k*5
a = [k]*5
for i in range(rest):
a[i+1]+=1
k = int(m/5)
rest = m - k*5
b = [k]*5
for i in range(rest):
b[i+1]+=1
r = a[0]*b[0] + a[1]*b[4] + a[2]*b[3]+ a[3]*b[2] + a[4]*b[1]
print(r)
main()
``` | instruction | 0 | 46,731 | 22 | 93,462 |
Yes | output | 1 | 46,731 | 22 | 93,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
"""
Author - Satwik Tiwari .
4th Oct , 2020 - Sunday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil
from copy import deepcopy
from collections import deque
# from collections import Counter as counter # Counter(list) return a dict with {key: count}
# from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
# from itertools import permutations as permutate
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 10**9+7
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve(case):
n,m = sep()
cnt1 = [0]*5
cnt2 = [0]*5
for i in range(1,m+1):
cnt1[i%5]+=1
for i in range(1,n+1):
cnt2[i%5]+=1
ans = 0
ans+=cnt1[0]*cnt2[0]
ans+=cnt1[1]*cnt2[4]
ans+=cnt1[2]*cnt2[3]
ans+=cnt1[3]*cnt2[2]
ans+=cnt1[4]*cnt2[1]
print(ans)
testcase(1)
# testcase(int(inp()))
``` | instruction | 0 | 46,732 | 22 | 93,464 |
Yes | output | 1 | 46,732 | 22 | 93,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
n, m = map(int, input().split())
q1 = n // 5
q2 = m // 5
res = 5 * q1 * q2
r1 = n % 5
r2 = m % 5
arr1 = [(i + 1) for i in range(r1)]
arr2 = [(i + 1) for i in range(r2)]
for i in range(len(arr1)):
for j in range(len(arr2)):
print(arr1[i], arr2[j])
if arr1[i] + arr2[j] % 5 == 0:
res += 1
print(i, j)
print(arr1, arr2)
res += r1 * q2
res += r2 * q1
print(res)
``` | instruction | 0 | 46,733 | 22 | 93,466 |
No | output | 1 | 46,733 | 22 | 93,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
n, m = tuple(input().split())
count = 0
for i in range(1, int(n)+1):
for j in range(1, int(m)+1):
if (i+j)%5 == 0:
break
count += (int(m)-j)//5 + 1
print(count)
``` | instruction | 0 | 46,734 | 22 | 93,468 |
No | output | 1 | 46,734 | 22 | 93,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
n , m = map(int,input().split())
def Alyona_Numbers(n,m):
if n%5 + m%5 >= 5 :return int(n*m/5)+1
else:return int(n*m/5)
print(Alyona_Numbers(n,m))
``` | instruction | 0 | 46,735 | 22 | 93,470 |
No | output | 1 | 46,735 | 22 | 93,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.
As usual, Alyona has some troubles and asks you to help.
Input
The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).
Output
Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.
Examples
Input
6 12
Output
14
Input
11 14
Output
31
Input
1 5
Output
1
Input
3 8
Output
5
Input
5 7
Output
7
Input
21 21
Output
88
Note
Following pairs are suitable in the first sample case:
* for x = 1 fits y equal to 4 or 9;
* for x = 2 fits y equal to 3 or 8;
* for x = 3 fits y equal to 2, 7 or 12;
* for x = 4 fits y equal to 1, 6 or 11;
* for x = 5 fits y equal to 5 or 10;
* for x = 6 fits y equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case.
Submitted Solution:
```
a,b=map(int,input().split(" "))
m=min(a,b)
n=max(a,b)
c=0
for i in range(1,m+1):
num=i%5
if num==0:
num=5
init=abs(5-num)
ans=(abs(n-init))//5
if ans!=0:
ans+=1
c+=ans
print(c)
``` | instruction | 0 | 46,736 | 22 | 93,472 |
No | output | 1 | 46,736 | 22 | 93,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,269 | 22 | 94,538 |
Tags: math, number theory
Correct Solution:
```
import math
n = int(input())
b = []
for i in range(n):
b.append(int(input()))
def lcm(a, b):
return abs(a*b) // math.gcd(a, b)
def sana(x):
i = 2
d = 1
b = 0
y = x
if x==1:
return 2
elif x==2:
return 5
else:
while(y!=0):
if d%i != 0:
d = lcm(d,i)
a = x//d
b = b + (y-a)*i
y = a
i+=1
if(len(str(b)) >= 10):
b = b%(10**9+7)
return b
for i in b:
print(sana(i))
``` | output | 1 | 47,269 | 22 | 94,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,270 | 22 | 94,540 |
Tags: math, number theory
Correct Solution:
```
#!/usr/bin/env pypy
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
from math import gcd
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
MOD = 10**9 + 7
# 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
def main():
for _ in range(int(input())):
n = int(input())
i = 2
x = 2
y = n
ans = 0
while y > 0:
x = (x // gcd(x, i)) * i
# ~ print(x, 2)
p = n // x
ans += (y - p) * i
y = p
i += 1
ans %= MOD
print(ans%MOD)
if __name__ == '__main__':
main()
``` | output | 1 | 47,270 | 22 | 94,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,271 | 22 | 94,542 |
Tags: math, number theory
Correct Solution:
```
import sys
import os.path
from collections import *
import math
import bisect
import heapq as hq
from fractions import Fraction
if (os.path.exists('input.txt')):
sys.stdin = open("input.txt", "r")
sys.stdout = open("output.txt", "w")
##########################################################
def lcm(x, y):
return x * y // math.gcd(x, y)
arr = [2]
n = 1
i = 2
h = Counter()
h[1] = 2
p = 10 ** 16 + 7
mod = 10 ** 9 + 7
arr = [1]
while n <= p:
x = lcm(n, i)
h[x] = (h[n] * (x // n) + 1) % mod
i += 1
n = x
arr.append(x)
arr.sort()
t = int(input())
while t:
res = 0
t -= 1
n = int(input())
while(n > 0):
x = 1
for i in range(len(arr)):
if arr[i] > n:
break
x = arr[i]
i = n // x
res = (res + (i * h[x])% mod) % mod
n -= i * x
print(res)
##########################################################
``` | output | 1 | 47,271 | 22 | 94,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,272 | 22 | 94,544 |
Tags: math, number theory
Correct Solution:
```
# import math
#
# for _ in range(int(input())):
# n, a, b = [int(i) for i in input().split()]
# m = b % a
# while True:
# if n <= 0:
# print('NO')
# break
# if n == 1:
# print('YES')
# break
# elif n % b == 1:
# print('YES')
# break
# elif n % a == 0:
# n //= a
# else:
# tmp = n % a
# sub = m * tmp // math.gcd(m, tmp)
# if sub == 0:
# print('NO')
# break
# n -= sub
def f(n):
for i in range(1, 10000):
if n % i != 0:
return i
import math
badguy = [[1, 2]]
for _ in range(30):
a, b = badguy[-1]
nxt = a * b // math.gcd(a, b)
badguy.append([nxt, f(nxt)])
for _ in range(int(input())):
n = int(input())
ret = 2 * n
pre = 2
for a, b in badguy[1:]:
m = n // a
ret -= m * pre
ret += m * b
pre = b
print(ret % int(1e9 + 7))
``` | output | 1 | 47,272 | 22 | 94,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,273 | 22 | 94,546 |
Tags: math, number theory
Correct Solution:
```
import sys
from math import gcd
def lcm(a,b):
return a*b//gcd(a,b)
input=sys.stdin.readline
t=int(input())
mod=10**9+7
f=[0]*45
f[1]=1
for i in range(2,45):
f[i]=lcm(f[i-1],i)
for _ in range(t):
n=int(input())
ans=0
for i in range(2,45):
r=n//f[i-1]-n//f[i]
ans=(ans+r*i%mod)%mod
print(ans)
``` | output | 1 | 47,273 | 22 | 94,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,274 | 22 | 94,548 |
Tags: math, number theory
Correct Solution:
```
from math import *
#from bisect import *
#from collections import *
#from random import *
#from decimal import *"""
#from heapq import *
#from random import *
import sys
input=sys.stdin.readline
sys.setrecursionlimit(3*(10**5))
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
p=10**9 + 7
while(t):
t-=1
n=inp()
r=(n+1)//2
r=r*2
r=r%p
ha=2
po=3
while(ha<=n):
q1=n//ha
ha1=(po*ha)//(gcd(po,ha))
q2=n//ha1
diff=q1-q2
ex=diff*po
po+=1
ha=ha1
r+=ex
r%=p
print(r)
``` | output | 1 | 47,274 | 22 | 94,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,275 | 22 | 94,550 |
Tags: math, number theory
Correct Solution:
```
import math
from sys import stdin, stdout
MOD = 10**9 + 7
dp = [1, 1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 12252240, 232792560, 232792560, 232792560, 232792560, 5354228880, 5354228880, 26771144400, 26771144400, 80313433200, 80313433200, 2329089562800, 2329089562800, 72201776446800, 144403552893600, 144403552893600,144403552893600,144403552893600,144403552893600,5342931457063200,5342931457063200,5342931457063200,5342931457063200,219060189739591200]
def solve():
t = int(input())
while t:
t -= 1
n = int(input())
x, a, b = 0, 0, 0
for i in range(41, 0, -1):
a = n // dp[i];
#print((i, n, dp[i], a, a*dp[i]))
#print((i+1), (a-b))
x += (a-b) * (i+1);
x %= MOD;
b = a;
print(x)
solve()
``` | output | 1 | 47,275 | 22 | 94,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10. | instruction | 0 | 47,276 | 22 | 94,552 |
Tags: math, number theory
Correct Solution:
```
import math
for _ in range(int(input())):
n = int(input())
s = 1
i = 2
ans = n
while s <= n:
ans += n//s
s = i*s//math.gcd(i,s)
i += 1
#print(s,i)
print(ans%(10**9+7))
``` | output | 1 | 47,276 | 22 | 94,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import sys, os
from io import BytesIO, IOBase
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
# 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")
stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
S1 = 'abcdefghijklmnopqrstuvwxyz'
S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
def lcm(a, b):
return a*b//gcd(a, b)
for _ in range(iinp()):
n = iinp()
ans = (2*n)%mod
x = 2
i = 3
while x <= n:
ans = (ans+n//x)%mod
x = lcm(x, i)
i += 1
print(ans)
``` | instruction | 0 | 47,277 | 22 | 94,554 |
Yes | output | 1 | 47,277 | 22 | 94,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_print = print
BUFSIZE = 8192
def dbg(*args, **kwargs):
_print('\33[95m', end='')
_print(*args, **kwargs)
_print('\33[0m', end='')
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')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
lcm = lambda a, b: a * b // gcd(a, b)
# ############################## main
MOD = 10 ** 9 + 7
MAX = 10 ** 16
A051451 = [1, 2, 6]
_idx = 3
while A051451[-1] < MAX:
_idx += 1
A051451.append(lcm(A051451[-1], _idx))
# dbg(A051451)
# dbg(len(A051451)) # 21
TABLE = tuple(enumerate(A051451, 2))[::-1]
# dbg(TABLE)
def solve():
n = itg()
ans = prev_count = 0
for f_i, a in TABLE:
count = n // a
ans += f_i * (count - prev_count)
prev_count = count
ans %= MOD
return ans
def main():
# print(solve())
for _ in range(itg()):
print(solve())
# solve()
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
DEBUG = 0
URL = 'https://codeforces.com/contest/1542/problem/C'
if __name__ == '__main__':
# 0: normal, 1: runner, 2: debug, 3: interactive
if DEBUG == 1:
import requests
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
if DEBUG != 2:
dbg = lambda *args, **kwargs: ...
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if DEBUG == 3:
def print(*args, **kwargs):
_print(*args, **kwargs)
sys.stdout.flush()
main()
# Please check!
``` | instruction | 0 | 47,278 | 22 | 94,556 |
Yes | output | 1 | 47,278 | 22 | 94,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import io,os,sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def readint():
return int(input().decode())
def lcm(x,y):
x,y = [max(x,y), min(x,y)]
if x%y == 0:
return x
m = lcm(y,x%y)
return x*(m//(x%y))
lcms = [0 for _ in range(150)]
lcms[1] = 1
for i in range(2,101):
lcms[i] = lcm(lcms[i-1], i)
for i in range(101,150):
lcms[i] = lcm(lcms[i-1], i)
if lcms[i] > int(1e17):
break
t = readint()
for _ in range(t):
n = readint()
if n== 1:
sys.stdout.write('2\n')
continue
ans = 0
i = 2
while lcms[i-1] <= n:
ans += i*(n//lcms[i-1] - n//lcms[i])
i += 1
sys.stdout.write('{}\n'.format(str(ans%int(1e9 + 7))))
``` | instruction | 0 | 47,279 | 22 | 94,558 |
Yes | output | 1 | 47,279 | 22 | 94,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import math
t=int(input())
while t:
t-=1
n=int(input())
res=2*n
a=2
b=3
while a<=n :
res+=(n//a)
res%=1000000007
a=(a*b)//math.gcd(a,b)
b+=1
print(res)
``` | instruction | 0 | 47,280 | 22 | 94,560 |
Yes | output | 1 | 47,280 | 22 | 94,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import math
for _ in range(int(input())):
n=int(input())
s=0
x=2
m=1
g=0
prev=1
MOD=1e9+7
left=n
N=n
for i in range(1,1000000000):
if(left<=0):
break
m=prev*i
g=math.gcd(prev,i)
prev=m//g
s=(s+i*(left-(N//prev))+MOD)%MOD
left=N//prev
print(int(s%MOD))
``` | instruction | 0 | 47,281 | 22 | 94,562 |
No | output | 1 | 47,281 | 22 | 94,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
import math;import heapq;import sys;input=sys.stdin.readline;S=lambda:input();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());H=1e9+7
lc=1
l=[]
t=pow(10,16)
i=1
while lc<=t:
lc=(lc*i)//math.gcd(lc,i)
l.append(lc)
i+=1
for _ in range(I()):
n=I()
s=(2*(n%H))%H
for i in range(1,len(l)):
if l[i]>n:
break
q=(n//l[i])%H
s=(s+q)%H
print(int(s))
``` | instruction | 0 | 47,282 | 22 | 94,564 |
No | output | 1 | 47,282 | 22 | 94,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
from sys import stdin
from math import floor, fmod
inp = stdin.readline
t = int(inp())
for _ in range(t):
n = int(inp())
i = 2
plus = 2*n
x = 2
while True:
if fmod(x, i) != 0:
for j in range(2, i+1):
if fmod(i, j) == 0:
x *= j
break
if x <= n:
plus += floor(n / int(x))
else:
break
plus = fmod(plus, (10**9 + 7))
i += 1
print(int(plus))
``` | instruction | 0 | 47,283 | 22 | 94,566 |
No | output | 1 | 47,283 | 22 | 94,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f(i) denote the minimum positive integer x such that x is not a divisor of i.
Compute ∑_{i=1}^n f(i) modulo 10^9+7. In other words, compute f(1)+f(2)+...+f(n) modulo 10^9+7.
Input
The first line contains a single integer t (1≤ t≤ 10^4), the number of test cases. Then t cases follow.
The only line of each test case contains a single integer n (1≤ n≤ 10^{16}).
Output
For each test case, output a single integer ans, where ans=∑_{i=1}^n f(i) modulo 10^9+7.
Example
Input
6
1
2
3
4
10
10000000000000000
Output
2
5
7
10
26
366580019
Note
In the fourth test case n=4, so ans=f(1)+f(2)+f(3)+f(4).
* 1 is a divisor of 1 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 1. Thus, f(1)=2.
* 1 and 2 are divisors of 2 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 2. Thus, f(2)=3.
* 1 is a divisor of 3 but 2 isn't, so 2 is the minimum positive integer that isn't a divisor of 3. Thus, f(3)=2.
* 1 and 2 are divisors of 4 but 3 isn't, so 3 is the minimum positive integer that isn't a divisor of 4. Thus, f(4)=3.
Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.
Submitted Solution:
```
def pr(k):
if k==1:
return 0
elif k==2:
return 1
else:
if k%2==0:
return 0
else:
d=3
while d*d<=k:
if k%d==0:
return 0
break
d+=2
else:
return 1
def h(p,n):
o=p**(n-1)
x=p**n
for i in range (1,x+1):
if pr(i)==1 and i!=p:
b=i
a=i
while a<p**n:
a=a*b
c=a//b
o*=c
return o
t=int(input())
for j in range (t):
n=int(input())
s=[2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41,43]
u=[h(s[i],1) for i in range (21)]
v=[n//u[i]-n//(u[i]*s[i]) for i in range (21)]
z=[v[i]*s[i] for i in range (21)]
print(sum(z)%(10**9+7))
``` | instruction | 0 | 47,284 | 22 | 94,568 |
No | output | 1 | 47,284 | 22 | 94,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,317 | 22 | 94,634 |
Tags: brute force
Correct Solution:
```
from math import sqrt
n,m=list(map(int,input().split()))
c=0
for i in range(int(sqrt(n))+1):
if (n-i**2)**2 + i == m:
c+=1
print(c)
``` | output | 1 | 47,317 | 22 | 94,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,318 | 22 | 94,636 |
Tags: brute force
Correct Solution:
```
import math
l = list(map(int,input().split()))
n,m = l[0],l[1]
A = min(m,int(math.sqrt(n)))
c = 0
for a in range(A+1):
if a+(n-a**2)**2 == m:
c = c+1
print(c)
``` | output | 1 | 47,318 | 22 | 94,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,319 | 22 | 94,638 |
Tags: brute force
Correct Solution:
```
c = 0
n, m = map(int,input().split())
for a in range(32):
for b in range(32):
if a*a+b == n and a+b*b == m:
c += 1
print(c)
``` | output | 1 | 47,319 | 22 | 94,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,320 | 22 | 94,640 |
Tags: brute force
Correct Solution:
```
import math
n, m = map(int, input().split())
ris = 0
for b in range(0, min([int(math.sqrt(m))+1, n+1])):
if b**4-2*m*b**2+b+m**2-n == 0:
tmp = n-b
if tmp >= 0:
tmp = math.sqrt(tmp)
if tmp == int(tmp):
ris +=1
print(ris)
``` | output | 1 | 47,320 | 22 | 94,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,321 | 22 | 94,642 |
Tags: brute force
Correct Solution:
```
import math
km = input()
count = 0
z = km.split()
n = int(z[0])
m = int(z[1])
a = int(math.sqrt(n))
b = 0
if(a*a != n):
b = n - (a*a)
if(a + (b*b) == m):
count += 1
bb = int(math.sqrt(m))
aa = 0
if(bb*bb != m):
aa = m - (bb*bb)
if((aa*aa) + bb == n):
if(aa == a and bb == b):
count = 1
elif(aa != a and bb != b):
count += 1
print(count)
``` | output | 1 | 47,321 | 22 | 94,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,322 | 22 | 94,644 |
Tags: brute force
Correct Solution:
```
n,m = map(int,input().split())
v,z =[],0
for i in range(m+1):
if i*i>m:break
a = m-i*i
v.append([a,i])
for i in v:
if i[0]**2 + i[1] ==n:z+=1
print(z)
``` | output | 1 | 47,322 | 22 | 94,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,323 | 22 | 94,646 |
Tags: brute force
Correct Solution:
```
n, m = map(int, input().split())
count = 0
k=int((n+m)**(1/2)) + 2
for a in range(0, k):
for b in range(0, k):
if a**2+b==n and b**2+a==m:
count+=1
print(count)
``` | output | 1 | 47,323 | 22 | 94,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
<image>
You should count, how many there are pairs of integers (a, b) (0 ≤ a, b) which satisfy the system.
Input
A single line contains two integers n, m (1 ≤ n, m ≤ 1000) — the parameters of the system. The numbers on the line are separated by a space.
Output
On a single line print the answer to the problem.
Examples
Input
9 3
Output
1
Input
14 28
Output
1
Input
4 20
Output
0
Note
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | instruction | 0 | 47,324 | 22 | 94,648 |
Tags: brute force
Correct Solution:
```
n,m = map(int,input().split())
cnt = 0
for i in range(1001):
for j in range(1001):
if i**2+j==n and i+j**2==m:
cnt+=1
print(cnt)
``` | output | 1 | 47,324 | 22 | 94,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,333 | 22 | 94,666 |
Tags: constructive algorithms, math
Correct Solution:
```
MOD = 10**9 + 9
n,m = map(int, input().split())
p = pow(2,m,MOD)
ans = 1
for i in range(1,n+1): ans = (ans * (p-i)) % MOD
print(ans)
``` | output | 1 | 47,333 | 22 | 94,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,334 | 22 | 94,668 |
Tags: constructive algorithms, math
Correct Solution:
```
a,b=map(int,input().split())
ans=1;mod=1000000009;gh=pow(2,b,mod)
for i in range(1,1+a):ans=(ans*(gh-i))%mod
print(ans)
``` | output | 1 | 47,334 | 22 | 94,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,335 | 22 | 94,670 |
Tags: constructive algorithms, math
Correct Solution:
```
n,m=map(int,input().split());MOD=1000000009;o=1;m=pow(2,m,MOD)-1
for i in range(n):o=o*(m-i)%MOD
print(o)
# Made By Mostafa_Khaled
``` | output | 1 | 47,335 | 22 | 94,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,336 | 22 | 94,672 |
Tags: constructive algorithms, math
Correct Solution:
```
n, m = map(int, input().split())
MOD = 10 ** 9 + 9
ans = pow(2, m, MOD) - 1
step = pow(2, m, MOD) - 2
for i in range(n - 1):
ans = (ans * step) % MOD
step -= 1
while ans < 0:
ans += MOD
while ans >= MOD:
ans -= MOD
print(ans)
``` | output | 1 | 47,336 | 22 | 94,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,338 | 22 | 94,676 |
Tags: constructive algorithms, math
Correct Solution:
```
MOD = int(1e9+9)
def fast_power(b, e):
res = 1
while e:
if e % 2 == 1:
res = res * b % MOD
b = b * b % MOD
e >>= 1
return res
n, m = map(int, input().split())
a = fast_power(2, m) - 1
a = (a + MOD) % MOD
b = a - 1
b = (b + MOD) % MOD
for i in range(1, n):
a = a * b % MOD
b = (b - 1 + MOD) % MOD
print(a)
``` | output | 1 | 47,338 | 22 | 94,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,339 | 22 | 94,678 |
Tags: constructive algorithms, math
Correct Solution:
```
n, m = map(int, input().split())
mod = 1000000009
upper = pow(2, m, mod)
s = 1
for i in range(0, n):
s = (s * (upper - i - 1)) % mod
print(s)
``` | output | 1 | 47,339 | 22 | 94,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r (1 ≤ l ≤ r ≤ n) such that <image>. In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).
Input
The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).
Output
Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.
Examples
Input
3 2
Output
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3). | instruction | 0 | 47,340 | 22 | 94,680 |
Tags: constructive algorithms, math
Correct Solution:
```
n, m = map(int, input().split())
MOD = 10 ** 9 + 9
out = 1
curr = pow(2, m, MOD) - 1
for i in range(n):
out *= curr
curr -= 1
out %= MOD
print(out)
``` | output | 1 | 47,340 | 22 | 94,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,534 | 22 | 95,068 |
Tags: implementation, number theory
Correct Solution:
```
p1, p2, p3, p4, a, b = map(int, input().split())
m = min(p1, p2, p3, p4)
if m > b:
print(b - a + 1)
else:
print(m - a if m - a > 0 else 0)
``` | output | 1 | 47,534 | 22 | 95,069 |
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