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'I' の数 (paizaランク B 相当)

Posted at

やってることは累積和の差し引きをして、その数を比較するだけのもの。
もうすこし累積和の特徴を知っておかなくては。。。

N,K = map(int,input().split())
A = [int(input()) for _ in range(N)]
# 累積和を求める
ans = [0] + A[:]
for i in range(1,N+1):
    ans[i] += ans[i-1]

for _ in range(K):
    a_left,a_right,b_left,b_right = map(int,input().split())

    # 累積和の足し引きで可能。ただしleftは−1引く必要がある
    a = ans[a_right] - ans[a_left - 1]
    b = ans[b_right] - ans[b_left - 1]
    
    #ルール確認・相手も反則してなければ必ず負けるようにする
    if a_right - a_left + 1 >= N/3:
        a = -1 
    if b_right - b_left + 1 >= N/3:
        b = -1 
 
    #判定
    if a == b:
        print("DRAW")
    elif a > b:
        print("A")
    else:
        print("B")

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