Problem:
 Score : 200 points    Problem Statement Takahashi has many red balls and blue balls. Now, he will place them in a row.  Initially, there is no ball placed.  Takahashi, who is very patient, will do the following operation 10^{100} times:   Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.   How many blue balls will be there among the first N balls in the row of balls made this way?      Constraints  1 \leq N \leq 10^{18}  A, B \geq 0  0 < A + B \leq 10^{18}  All values in input are integers.         Input Input is given from Standard Input in the following format:  N  A  B       Output Print the number of blue balls that will be there among the first N balls in the row of balls.        Sample Input 1 8 3 4      Sample Output 1 4  Let b denote a blue ball, and r denote a red ball. The first eight balls in the row will be bbbrrrrb , among which there are four blue balls.       Sample Input 2 8 0 4      Sample Output 2 0  He placed only red balls from the beginning.       Sample Input 3 6 2 4      Sample Output 3 2  Among bbrrrr , there are two blue balls.

Solution (Python):

from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    n,a,b = LI()
    s = a+b
    k = n//s
    n -= k*s
    n = min(n,a)
    print(k*a+n)
    return


if __name__ == "__main__":
    solve()

Solution (Python):
n,a,b=map(int,input().split())
l=n//(a+b)
r=n%(a+b)
print(l*a+min(r,a))

Solution (Python):
import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def main():
    n,a,b=MI()
    p,r=divmod(n,(a+b))
    print(p*a+min(r,a))

main()

Solution (Python):
n, a, b = map(int, input().split())
print(n // (a + b) * a + min(n % (a + b), a))

Solution (Python):
N, A, B = map(int, input().split())
ans = N // (A + B) * A
rem = N % (A + B)
ans += min(rem, A)
print(ans)