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Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,284
8
186,568
"Correct Solution: ``` n = int(input()) xl = [] for i in range(n): x,l = map(int,input().split()) xl.append([x+l,x-l]) xl.sort() ans = -10**10 count = 0 for j in range(n): if xl[j][1] >= ans: count +=1 ans = xl[j][0] print(count) ```
output
1
93,284
8
186,569
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,285
8
186,570
"Correct Solution: ``` n=int(input()) xls=[list(map(int,input().split())) for _ in range(n)] lrs=sorted((x-l,x+l) for x,l in xls) res=0 most_r=lrs[0][0] for l,r in lrs: if l>=most_r: res+=1 most_r=r elif most_r>r: most_r=r print(res) ```
output
1
93,285
8
186,571
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,286
8
186,572
"Correct Solution: ``` n = int(input()) count = 0 X = [] for i in range(n): x,l = map(int, input().split()) X.append((x+l,x-l)) X.sort() coord = -float('inf') for i in X: if i[1] >= coord: count += 1 coord = i[0] print(count) ```
output
1
93,286
8
186,573
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,287
8
186,574
"Correct Solution: ``` n=int(input()) XL=[] for i in range(n): x,l=map(int,input().split()) XL.append([x-l,x,x+l]) ans=0 cur=-float("inf") XL.sort(key=lambda x:x[2]) for x1,x2,x3 in XL: if x1>=cur: cur=x3 ans+=1 else: continue print(ans) ```
output
1
93,287
8
186,575
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,288
8
186,576
"Correct Solution: ``` N = int(input()) arms = [] for i in range(0, N): X, L = map(int, input().split(" ")) arms.append((X + L, X - L)) arms.sort() ans = 0 coord = -float('inf') for arm in arms: if arm[1] >= coord: ans += 1 coord = arm[0] print(ans) ```
output
1
93,288
8
186,577
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,289
8
186,578
"Correct Solution: ``` N = int(input()) P = [] for _ in range(N): x,l = list(map(int, input().split())) P.append((x+l, x-l)) P.sort() rw = P[0][0] ans = 1 for i in range(1,N): r,l = P[i] if l < rw: continue else: rw = r ans += 1 print(ans) ```
output
1
93,289
8
186,579
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,290
8
186,580
"Correct Solution: ``` N = int(input()) q = [] for i in range(N): x,l = map(int, input().split()) q.append((x+l, x-l)) q.sort() cnt = 0 INF = float("inf") b = -INF for r,l in q: if l >= b: b = r else: cnt += 1 print(N-cnt) ```
output
1
93,290
8
186,581
Provide a correct Python 3 solution for this coding contest problem. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5
instruction
0
93,291
8
186,582
"Correct Solution: ``` n=int(input()) d=[] for i in range(n): x,l=map(int,input().split()) d.append((x-l,x+l)) d.sort() ans=1 chk=d[0][1] for i in d: if i[0]>=chk: ans+=1 chk=i[1] else: chk=min(chk,i[1]) print(ans) ```
output
1
93,291
8
186,583
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` n=int(input()) XL=[list(map(int,input().split())) for i in range(n)] Range=[[x-l,x+l] for x,l in XL] ans,end=0,-1*(10**9+1) Range.sort(key=lambda X: X[1]) for R in Range: if end > R[0]: continue ans += 1 end=R[1] print(ans) ```
instruction
0
93,292
8
186,584
Yes
output
1
93,292
8
186,585
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` N, *XL = map(int, open(0).read().split()) XL = sorted((x + l, x - l) for x, l in zip(*[iter(XL)] * 2)) curr = -float("inf") ans = 0 for right, left in XL: if left >= curr: curr = right ans += 1 print(ans) ```
instruction
0
93,293
8
186,586
Yes
output
1
93,293
8
186,587
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` n=int(input()) L=[] for i in range(n): x,l=map(int,input().split()) L.append([x-l,x+l]) L.sort(key=lambda x:x[1]) ans=1 bef=L[0][1] for i in range(1,n): if bef<=L[i][0]: bef=L[i][1] ans+=1 print(ans) ```
instruction
0
93,294
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186,588
Yes
output
1
93,294
8
186,589
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` n=int(input()) l=[] for i in range(n): x,y=map(int,input().split()) l.append([x-y,x+y]) count=0 l=sorted(l) l=l[::-1] count+=1 now=l[0][0] for i in range(1,n): if now>=l[i][1]: now=l[i][0] count+=1 else: pass print(count) ```
instruction
0
93,295
8
186,590
Yes
output
1
93,295
8
186,591
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` N=int(input()) X=[] L=[] XA=[] XB=[] Q=[0]*N for i in range(N): x,l=map(int,input().split()) X.append(x) L.append(l) for i in range(N): XA.append(X[i]-L[i]) XB.append(X[i]+L[i]) S=sorted(list(zip(XA,XB))) cnt=0 #print(S) i=0 while i<N-1: if i==0: if S[i][1]>S[i+1][0]: cnt=cnt+1 if S[i][1]<=S[i+1][1]: S.pop(i+1) else: S.pop(i) i=i+1 else: i=i+1 else: if S[i-1][1]>S[i][0]: cnt=cnt+1 S.pop(i) i=i+1 else: i=i+1 #print(cnt) #print(S) print(N-cnt) ```
instruction
0
93,296
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No
output
1
93,296
8
186,593
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` n = int(input()) xl = [list(map(int, input().split())) for _ in range(n)] xx = [] count = 0 for x in xl: xx.append([x[0] - x[1], x[0] + x[1]]) if n != 2: for i in range(1, n-1): if xx[i][0] < xx[i-1][1] and xx[i][1] > xx[i+1][0]: count += 1 print(n-count) else: if xx[1][0] < xx[0][1]: print(1) else: print(2) ```
instruction
0
93,297
8
186,594
No
output
1
93,297
8
186,595
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` import sys from pprint import pprint def solve(n, x, l): r = [] for i in range(n): r.append((x[i], l[i])) x_pre = -1 l_pre = 0 remove = 0 for x_i, l_i in sorted(r): if x_pre == -1: x_pre = x_i l_pre = l_i continue # print("pre", x_pre, l_pre) # print(x_i, l_i) if x_i - l_i < x_pre + l_pre: # 先まで届かない方を残す if x_pre + l_pre > x_i + l_i: x_pre = x_i l_pre = x_i remove += 1 print(n - remove) if __name__ == '__main__': n = int(sys.stdin.readline().strip()) x = [0] * n l = [0] * n for i in range(n): x[i], l[i] = map(int, sys.stdin.readline().strip().split(" ")) solve(n, x, l) ```
instruction
0
93,298
8
186,596
No
output
1
93,298
8
186,597
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep. Constraints * 1 \leq N \leq 100,000 * 0 \leq X_i \leq 10^9 (1 \leq i \leq N) * 1 \leq L_i \leq 10^9 (1 \leq i \leq N) * If i \neq j, X_i \neq X_j. * All values in input are integers. Input Input is given from Standard Input in the following format: N X_1 L_1 X_2 L_2 \vdots X_N L_N Output Print the maximum number of robots that we can keep. Examples Input 4 2 4 4 3 9 3 100 5 Output 3 Input 2 8 20 1 10 Output 1 Input 5 10 1 2 1 4 1 6 1 8 1 Output 5 Submitted Solution: ``` n = int(input()) x = [] for i in range(n): x1, y1=[int(i) for i in input().split()] x.append([x1, y1]) x.sort() remove = 0 stocks = 0 for k in range(n - 1): if k+1+stocks < n: if (x[k + 1 + stocks][0] - x[k][0]) < x[k][1] + x[k + 1][1]: remove += 1 x[k + 1][0] = -1 if n - 2 > k: stocks += 1 k -=1 else: stocks=0 print (n -remove) ```
instruction
0
93,299
8
186,598
No
output
1
93,299
8
186,599
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,548
8
187,096
Tags: data structures, implementation Correct Solution: ``` n = int(input()) a = list(map(int,input().split())) s = [] flag = 1 for i in range(0,n): if len(s) and s[-1] == a[i]: s.pop() elif len(s) == 0 or s[-1] > a[i]: s.append(a[i]) else: print("NO") flag = 0 break if(flag == 1): if len(s) == 0 or len(s) == 1 and s[-1] == max(a): print("YES") else: print("NO") ```
output
1
93,548
8
187,097
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,549
8
187,098
Tags: data structures, implementation Correct Solution: ``` n=int(input()) a=list(map(int,input().split())) down=[] good=True for guy in a: if len(down)==0: down.append(guy) elif down[-1]>guy: down.append(guy) elif down[-1]==guy: down.pop() else: good=False break if not good: print("NO") elif len(down)>1: print("NO") elif len(down)==1 and down[0]!=max(a): print("NO") else: print("YES") ```
output
1
93,549
8
187,099
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,550
8
187,100
Tags: data structures, implementation Correct Solution: ``` i,p,l,j=input,print,len,int n,m,q=j(i()),0,[] f=q.append for a in map(j,i().split()): if q: if a==q[-1]:q.pop() elif a>q[-1]:f(a);break else:f(a) else:f(a) m=max(m,a) if l(q)==0 or l(q)==1 and q[0]==m:p('YES') else:p('NO') ```
output
1
93,550
8
187,101
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,551
8
187,102
Tags: data structures, implementation Correct Solution: ``` n = int(input()) a = list(map(int, input().split())) b = [[a[0], 1]] fl = False for i in range(1, n): if fl: break if a[i] == b[-1][0]: b[-1][1] += 1 elif a[i] < b[-1][0]: b.append([a[i], 1]) else: while len(b) != 0 and a[i] > b[-1][0]: if b[-1][1] % 2 != 0: print("NO") fl = True break b.pop() if len(b) == 0 or a[i] < b[-1][0]: b.append([a[i], 1]) else: b[-1][1] += 1 if not fl: i = len(b) - 1 while i != 0: if b[i][1] % 2 != 0: fl = True break i -= 1 if fl: print("NO") else: print("YES") ```
output
1
93,551
8
187,103
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,552
8
187,104
Tags: data structures, implementation Correct Solution: ``` i = int(input()) s = input().split() m = max (list(map(int,s))) l = [] for j in s: if not l or int(j) < l[-1] : l.append(int(j)) elif int(j) == l[-1]: l.pop() else: print('NO') exit() if (len(l) ==1 and l[0] >= m) or len(l) == 0 : print('YES') else: print('NO') ```
output
1
93,552
8
187,105
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,553
8
187,106
Tags: data structures, implementation Correct Solution: ``` i,p=input,print;n,m,q=int(i()),0,[];f=q.append for a in map(int,i().split()): if q: if a==q[-1]:q.pop() elif a>q[-1]:f(a);break else:f(a) else:f(a) m=max(m,a) p('YES'if len(q)==0 or len(q)==1 and q[0]==m else'NO') ```
output
1
93,553
8
187,107
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,554
8
187,108
Tags: data structures, implementation Correct Solution: ``` n=int(input()) a=list(map(int,input().split())) b=[0]*1000000 sp=-1 max=0 f=True; for i in range(n): sp=sp+1 b[sp]=a[i] if b[sp]>max: max=b[sp] if sp>0: if b[sp]>b[sp-1]: f=False if b[sp]==b[sp-1]: sp=sp-2 if f==True : if sp==0: if b[sp]>=max: print("YES") else: print("NO") elif sp<0: print("YES") else: print("NO") else: print("NO") ```
output
1
93,554
8
187,109
Provide tags and a correct Python 3 solution for this coding contest problem. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete.
instruction
0
93,555
8
187,110
Tags: data structures, implementation Correct Solution: ``` n=int(input()) l=[int(x) for x in input().split()] st=[] for i in l: d=i if len(st) and st[-1]==d: st.pop() elif len(st)==0 or st[-1]>d: st.append(d) else: print("NO") break else: if len(st)==0 or len(st)==1 and st[-1]==max(l): print("YES") else: print("NO") ```
output
1
93,555
8
187,111
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` import sys n = int(sys.stdin.readline()) cur_str = ((sys.stdin.readline()).strip()).split() array = [int(elem) for elem in cur_str] i = 0 stack = [] ans = True maximum = -1 while i<len(array): maximum = max(maximum,array[i]) if not stack or array[i] < stack[-1]: stack.append(array[i]) else: if stack[-1] < array[i]: ans = False break elif stack[-1] == array[i]: stack.pop() i+=1 for i in range(1,len(stack)): if stack[i]!=stack[i-1]: ans = False break if stack and min(stack) < maximum: ans = False if ans == True or n==1: print("YES") else: print("NO") ```
instruction
0
93,556
8
187,112
Yes
output
1
93,556
8
187,113
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` from bisect import bisect_right as br from bisect import bisect_left as bl from collections import defaultdict from itertools import combinations import functools import sys import math MAX = sys.maxsize MAXN = 10**6+10 MOD = 998244353 def isprime(n): n = abs(int(n)) if n < 2: return False if n == 2: return True if not n & 1: return False for x in range(3, int(n**0.5) + 1, 2): if n % x == 0: return False return True def mhd(a,b,x,y): return abs(a-x)+abs(b-y) def numIN(): return(map(int,sys.stdin.readline().strip().split())) def charIN(): return(sys.stdin.readline().strip().split()) t = [(-1,-1)]*1000010 def create(a): global t,n for i in range(n,2*n): t[i] = (a[i-n],i-n) for i in range(n-1,0,-1): x = [t[2*i],t[2*i+1]] x.sort(key = lambda x:x[0]) t[i] = x[1] def update(idx,value): global t,n idx = idx+n t[idx] = value while(idx>1): idx = idx//2 x = [t[2*idx],t[2*idx+1]] x.sort(key = lambda x:x[0]) t[idx] = x[1] def dis(a,b,k): ans = 0 for i in range(k): ans+=abs(a[i]-b[i]) return ans def cal(n,k): res = 1 c = [0]*(k+1) c[0]=1 for i in range(1,n+1): for j in range(min(i,k),0,-1): c[j] = (c[j]+c[j-1])%MOD return c[k] n = int(input()) l = list(numIN()) x = [] for i in range(n): if x and x[-1]==l[i]: x.pop() continue if x and x[-1]<l[i]: print('NO') exit(0) x.append(l[i]) if x and x[-1]!=max(l): print('NO') else: print('YES') ```
instruction
0
93,557
8
187,114
Yes
output
1
93,557
8
187,115
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` import sys input = sys.stdin.readline from collections import deque n=int(input()) A=list(map(int,input().split())) AX=[(A[i],i) for i in range(n)] QUE = deque() for i in range(n): QUE.append(AX[i]) while len(QUE)>=2 and QUE[-1][0]==QUE[-2][0] and A[QUE[-2][1]]>=A[QUE[-2][1]+1]: QUE.pop() QUE.pop() if len(QUE)>=2: print("NO") elif len(QUE)==1: if QUE[0][0]==max(A): print("YES") else: print("NO") else: print("YES") ```
instruction
0
93,558
8
187,116
Yes
output
1
93,558
8
187,117
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` def main(): n = int(input()) m = 0 stack = [] for val in map(int, input().split()): if stack: if val == stack[-1]: stack.pop() elif val > stack[-1]: stack.append(val) break else: stack.append(val) else: stack.append(val) m = max(m, val) print('YES' if len(stack) == 0 or len(stack) == 1 and stack[0] == m else 'NO') if __name__ == '__main__': main() ```
instruction
0
93,559
8
187,118
Yes
output
1
93,559
8
187,119
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` n=int(input()) a=list(map(int,input().split())) b=[0]*10000 sp=0 for i in range(n): b[sp]=a[i] sp=sp+1 if sp>0: if b[sp-1]==b[sp-2]: sp=sp-2 if sp<=1: print("YES") else: print("NO") ```
instruction
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8
187,120
No
output
1
93,560
8
187,121
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` #!/usr/bin/env python3 import sys def rint(): return map(int, sys.stdin.readline().split()) #lines = stdin.readlines() n = int(input()) a = list(rint()) stack = [a[0]] for i in range(1, n): if len(stack) == 0: stack.append(a[i]) if a[i] == stack[-1]: stack.pop() else: stack.append(a[i]) if len(stack) > 1: print("NO") else: print("YES") ```
instruction
0
93,561
8
187,122
No
output
1
93,561
8
187,123
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` n=int(input()) l=[int(x) for x in input().split()] st=[] for i in l: d=i if len(st) and st[-1]==d: st.pop() else: st.append(d) if len(st)<2: print("YES") else: print("NO") ```
instruction
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93,562
8
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No
output
1
93,562
8
187,125
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The current state of the wall can be respresented by a sequence a of n integers, with a_i being the height of the i-th part of the wall. Vova can only use 2 × 1 bricks to put in the wall (he has infinite supply of them, however). Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some i the current height of part i is the same as for part i + 1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n of it). Note that Vova can't put bricks vertically. Vova is a perfectionist, so he considers the wall completed when: * all parts of the wall has the same height; * the wall has no empty spaces inside it. Can Vova complete the wall using any amount of bricks (possibly zero)? Input The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of parts in the wall. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial heights of the parts of the wall. Output Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero). Print "NO" otherwise. Examples Input 5 2 1 1 2 5 Output YES Input 3 4 5 3 Output NO Input 2 10 10 Output YES Note In the first example Vova can put a brick on parts 2 and 3 to make the wall [2, 2, 2, 2, 5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5, 5, 5, 5, 5]. In the second example Vova can put no bricks in the wall. In the third example the wall is already complete. Submitted Solution: ``` n=int(input()) a=list(map(int,input().split())) b=[0]*100001 sp=-1 max=0 f=True; for i in range(n): sp=sp+1 b[sp]=a[i] if b[sp]>max: max=b[sp] if sp>0: if b[sp]>b[sp-1]: f=False if b[sp]==b[sp-1]: sp=sp-2 if f==True : if sp==0: if b[sp]>max: print("YES") else: print("NO") elif sp<0: print("YES") else: print("NO") ```
instruction
0
93,563
8
187,126
No
output
1
93,563
8
187,127
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,612
8
187,224
Tags: math Correct Solution: ``` n=int(input()) for _ in range(n): c, s = map(int,input().split()) q = s//c r = s%c print(r*(q+1)**2 + (c-r)*q*q) ```
output
1
93,612
8
187,225
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,613
8
187,226
Tags: math Correct Solution: ``` t = int(input()) for c in range (t): ss = 0 a,b= map(int, input().split()) if b<=a: print(b) else : de = int(b/a) ss = de**2 *(a-b%a) +(de+1)**2 *(b%a) print(ss) ```
output
1
93,613
8
187,227
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,614
8
187,228
Tags: math Correct Solution: ``` ''' Hey stalker :) ''' INF = 10**10 def main(): print = out.append ''' Cook your dish here! ''' c, sum = get_list() x = 0 while (x)*c<=sum: x+=1 res = c*(x-1)**2 res += (sum - (x-1)*c)*(2*x-1) print(res) ''' Pythonista fLite 1.1 ''' import sys #from collections import defaultdict, Counter #from functools import reduce #import math input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__ out = [] get_int = lambda: int(input()) get_list = lambda: list(map(int, input().split())) #main() [main() for _ in range(int(input()))] print(*out, sep='\n') ```
output
1
93,614
8
187,229
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,615
8
187,230
Tags: math Correct Solution: ``` for i in range(int(input())): c, sumi = [int(x) for x in input().split()] deg = sumi//c res = ((deg + 1)**2) * (sumi%c) + (c - sumi%c)*(deg**2) print(res) ```
output
1
93,615
8
187,231
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,616
8
187,232
Tags: math Correct Solution: ``` n = int(input()) for i in range(n): c, s = map(int, input().strip().split()) left = s % c v = s // c o = 0 for i in range(left): o += (v + 1) ** 2 for i in range(left, c): o += v ** 2 print(o) ```
output
1
93,616
8
187,233
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,617
8
187,234
Tags: math Correct Solution: ``` n = int(input()) for _ in range(n): c, su = map(int, input().split()) ci = [0] * c while su > 0: for i in range(c): if su > 0: ci[i] += 1 su -= 1 else: break to = 0 for i in range(c): to += ci[i] ** 2 print(to) ```
output
1
93,617
8
187,235
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,618
8
187,236
Tags: math Correct Solution: ``` from collections import defaultdict t = int(input()) while(t>0): a,b= [int(i) for i in input().split()] if a==1 or b==1: ans = b**2 else: curr = b//a rem = b%a temp = curr-rem s = a-rem ans = s*(curr**2)+rem*(curr+1)**2 print(ans) t-=1 ```
output
1
93,618
8
187,237
Provide tags and a correct Python 3 solution for this coding contest problem. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18.
instruction
0
93,619
8
187,238
Tags: math Correct Solution: ``` ''' Created on 2019. 9. 21. @author: kkhh88 ''' #q = int(input()) #x, y = map(int,input().split(' ')) #print (' '.join(list(map(str, s)))) q = int(input()) for i in range(q): x, y = map(int,input().split(' ')) c = y // x d = y % x sum = (c + 1) * (c + 1) * d + c * c * (x - d) print (sum) ```
output
1
93,619
8
187,239
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` T = int(input()) anwer = "" for t in range(T): text = input().split(' ') maximum = int(text[0]) need = int(text[1]) one = need // maximum if one == 0: anwer += str(need) + '\n' continue ost = need - maximum * one if ost == 0: n = need / one anwer += str(int(one**2 * n)) + '\n' continue boost = ost // maximum boost2 = ost - maximum * boost one += boost cost = 0 k = maximum - boost2 for i in range(maximum): if boost2: cost += (one + 1)**2 boost2 -= 1 else: cost += one**2 * k break anwer += str(int(cost)) + '\n' print(anwer) ```
instruction
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` n = int(input()) for _ in range(n): c,k = map(int,input().strip().split()) x = k//c y = k%c ans = y*((x+1)**2)+(c-y)*(x**2) print(ans) ```
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Yes
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1
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` for i in ' '*int(input()): c,sum=map(int,input().split()) k=sum//c r=sum%c print((k+1)**2*r+k**2*(c-r)) ```
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Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` for i in range(int(input())): a, b = map(int, input().split()) if a == 1 or b == 1: print(pow(b, 2)) else: ans = [0]*a while(b != 0): for i in range(a): if b == 0: break else: ans[i] += 1 b -= 1 count = 0 for i in range(len(ans)): count += pow(ans[i], 2) print(count) ```
instruction
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93,623
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Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` n = int(input()) while(n): sum = 0 a,b = map(int,input().split()) if a == 1: print(b^2) elif a>b: print(b) else: z = (b//a) +1 x = z-1 m = b%a r = a - m sum += (m * (z^2)) sum += (r * (x^2)) print(sum) n-=1 ```
instruction
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` n = int(input()) for i in range(n): a, b = map(int,input().split()) count = 0 while (b%a!=0 and b!=0): count+=1 b-=1 a-=1 count+=((b//a)**2)*a print(count) ```
instruction
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` from math import gcd for _ in range(int(input())): ci,si=map(int,input().split()) if si<ci: print(si) else: ans=0 t=si//2 rem = si%2 if rem: ans+=t**2*(ci-1) ans+=(rem+t)**2 else: ans+=t**2*ci print(ans) ```
instruction
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Several days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house. Your house has n rooms. In the i-th room you can install at most c_i heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to k^2 burles. Since rooms can have different sizes, you calculated that you need at least sum_i sections in total in the i-th room. For each room calculate the minimum cost to install at most c_i radiators with total number of sections not less than sum_i. Input The first line contains single integer n (1 ≤ n ≤ 1000) — the number of rooms. Each of the next n lines contains the description of some room. The i-th line contains two integers c_i and sum_i (1 ≤ c_i, sum_i ≤ 10^4) — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively. Output For each room print one integer — the minimum possible cost to install at most c_i radiators with total number of sections not less than sum_i. Example Input 4 1 10000 10000 1 2 6 4 6 Output 100000000 1 18 10 Note In the first room, you can install only one radiator, so it's optimal to use the radiator with sum_1 sections. The cost of the radiator is equal to (10^4)^2 = 10^8. In the second room, you can install up to 10^4 radiators, but since you need only one section in total, it's optimal to buy one radiator with one section. In the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs 3^2+ 3^2 = 18. Submitted Solution: ``` def process(): a,b = input().split() a,b = int(a), int(b) if b == 3 and a == 2: print(5) return if a > b: print(1) else: if a == 1: print(int(b**2)) return if b % a == 0: res = 0 ave = int(b/a) while b != 0: res += ave**2 b -= ave print(res) else: res = 0 ave = int(b/a) tmp = b/a - ave cnt = 0 for i in range(1,a+1): if i*tmp == int(i*tmp): cnt = i break for i in range(cnt): res += int(ave)**2 for i in range(a-cnt): res += (int(ave)+1)**2 print(res) tests = int(input()) for test in range(tests): process() ```
instruction
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No
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Provide tags and a correct Python 3 solution for this coding contest problem. Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks. At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C. Given number n, find the minimally possible and maximally possible number of stolen hay blocks. Input The only line contains integer n from the problem's statement (1 ≤ n ≤ 109). Output Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves. Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator. Examples Input 4 Output 28 41 Input 7 Output 47 65 Input 12 Output 48 105 Note Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
instruction
0
93,683
8
187,366
Tags: brute force, math Correct Solution: ``` n = int(input()) u = v = 9 + 8 * n for a in range(1, int(n ** (1 / 3)) + 1): if n % a: continue m = n // a for b in range(a, int(m ** 0.5) + 1): if m % b: continue p = (a + 1) * (b + 2) * (m // b + 2) - n if u > p: u = p print(u, v) ```
output
1
93,683
8
187,367
Provide tags and a correct Python 3 solution for this coding contest problem. Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks. At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C. Given number n, find the minimally possible and maximally possible number of stolen hay blocks. Input The only line contains integer n from the problem's statement (1 ≤ n ≤ 109). Output Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves. Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator. Examples Input 4 Output 28 41 Input 7 Output 47 65 Input 12 Output 48 105 Note Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
instruction
0
93,684
8
187,368
Tags: brute force, math Correct Solution: ``` from math import sqrt p, n = [], int(input()) def f(x, y): return (x + 2) * (y + 2) + (2 * (x + y + 2) * n) // (x * y) for x in range(2, int(sqrt(n)) + 1): if n % x == 0: p.append(x) p += [n // x for x in reversed(p)] p.append(n) u = v = f(1, 1) for m in p: for x in range(1, int(sqrt(m)) + 1): if m % x == 0: u = min(u, f(x, m // x)) print(u, v) # Made By Mostafa_Khaled ```
output
1
93,684
8
187,369