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๋ฐฑ์ค€ - ๋™์ „ 0(๊ทธ๋ฆฌ๋”” ์•Œ๊ณ ๋ฆฌ์ฆ˜), ํŒŒ์ด์ฌ & C

by Janger 2021. 12. 5.
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https://www.acmicpc.net/problem/11047

 

11047๋ฒˆ: ๋™์ „ 0

์ฒซ์งธ ์ค„์— N๊ณผ K๊ฐ€ ์ฃผ์–ด์ง„๋‹ค. (1 ≤ N ≤ 10, 1 ≤ K ≤ 100,000,000) ๋‘˜์งธ ์ค„๋ถ€ํ„ฐ N๊ฐœ์˜ ์ค„์— ๋™์ „์˜ ๊ฐ€์น˜ Ai๊ฐ€ ์˜ค๋ฆ„์ฐจ์ˆœ์œผ๋กœ ์ฃผ์–ด์ง„๋‹ค. (1 ≤ Ai ≤ 1,000,000, A1 = 1, i ≥ 2์ธ ๊ฒฝ์šฐ์— Ai๋Š” Ai-1์˜ ๋ฐฐ์ˆ˜)

www.acmicpc.net

 

[ํŒŒ์ด์ฌ]

N, K = map(int, input().split())
	
coins = []
count = 0
for _ in range(N):
	coins.append( int( input() ) )

idx = len( coins ) - 1
	
while K > 0:

	if ( K - coins[idx] ) >= 0 :
		count += 1
		K -= coins[idx]
		idx = len(coins) - 1
		continue
		
	idx -= 1

		
print( count )

๊ทธ๋ƒฅ ์‹คํ–‰์‹œํ‚ค๋ฉด ์‹œ๊ฐ„ ์˜ค๋ฅ˜๊ฐ€ ๋œจ๊ธฐ ๋•Œ๋ฌธ์— PyPy3๋กœ ์‹คํ–‰์‹œ์ผœ๋ณด๋‹ˆ ์ž˜ ๋œ๋‹ค. 

 

 

[C]

#include <stdio.h>

int main(void)
{
	int N, K, arr[101];
	
	scanf("%d %d", &N, &K);
	
	int i=0;
	
	for(i=0; i<N; i++){
		scanf("%d", &arr[i]);
	}
	
	int count = 0;
	for(i=N-1; i>=0; i--){
		if( K / arr[i] >= 0 ){
			count += K / arr[i];
			K %= arr[i];
		}
	}
	
	printf("%d", count);
	
	return 0;
}

์–˜๋Š” ๋” ํšจ์œจ์ ์œผ๋กœ ๋‚˜๋จธ์ง€์™€ ๋ชซ์œผ๋กœ ๊ตฌํ•˜๋Š” ๋ฐฉ๋ฒ•

 

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