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[Silver II] Title: 이항 계수 2, Time: 260 ms, Memory: 55236 KB -BaekjoonHub

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# [Silver II] 이항 계수 2 - 11051
[문제 링크](https://www.acmicpc.net/problem/11051)
### 성능 요약
메모리: 55236 KB, 시간: 260 ms
### 분류
조합론, 다이나믹 프로그래밍, 수학
### 제출 일자
2025년 3월 4일 17:53:01
### 문제 설명
<p>자연수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(N\)</span></mjx-container>과 정수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(K\)</span></mjx-container>가 주어졌을 때 이항 계수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mrow><mjx-texatom texclass="OPEN"><mjx-mo class="mjx-sop"><mjx-c class="mjx-c28 TEX-S1"></mjx-c></mjx-mo></mjx-texatom><mjx-mfrac><mjx-frac atop="true" delims="true" style="vertical-align: -0.345em;"><mjx-num style="padding-bottom: 0.306em;"><mjx-mi class="mjx-i" size="s"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-num><mjx-den><mjx-mi class="mjx-i" size="s"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-den></mjx-frac></mjx-mfrac><mjx-texatom texclass="CLOSE"><mjx-mo class="mjx-sop"><mjx-c class="mjx-c29 TEX-S1"></mjx-c></mjx-mo></mjx-texatom></mjx-mrow></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow data-mjx-texclass="ORD"><mrow data-mjx-texclass="OPEN"><mo minsize="1.2em" maxsize="1.2em">(</mo></mrow><mfrac linethickness="0"><mi>N</mi><mi>K</mi></mfrac><mrow data-mjx-texclass="CLOSE"><mo minsize="1.2em" maxsize="1.2em">)</mo></mrow></mrow></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(\binom{N}{K}\)</span></mjx-container>를 10,007로 나눈 나머지를 구하는 프로그램을 작성하시오.</p>
### 입력
<p>첫째 줄에 <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(N\)</span></mjx-container>과 <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(K\)</span></mjx-container>가 주어진다. (1 ≤ <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(N\)</span></mjx-container> ≤ 1,000, 0 ≤ <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(K\)</span></mjx-container> ≤ <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(N\)</span></mjx-container>)</p>
### 출력
<p> <mjx-container class="MathJax" jax="CHTML" style="font-size: 109%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mrow><mjx-texatom texclass="OPEN"><mjx-mo class="mjx-sop"><mjx-c class="mjx-c28 TEX-S1"></mjx-c></mjx-mo></mjx-texatom><mjx-mfrac><mjx-frac atop="true" delims="true" style="vertical-align: -0.345em;"><mjx-num style="padding-bottom: 0.306em;"><mjx-mi class="mjx-i" size="s"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-num><mjx-den><mjx-mi class="mjx-i" size="s"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-den></mjx-frac></mjx-mfrac><mjx-texatom texclass="CLOSE"><mjx-mo class="mjx-sop"><mjx-c class="mjx-c29 TEX-S1"></mjx-c></mjx-mo></mjx-texatom></mjx-mrow></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow data-mjx-texclass="ORD"><mrow data-mjx-texclass="OPEN"><mo minsize="1.2em" maxsize="1.2em">(</mo></mrow><mfrac linethickness="0"><mi>N</mi><mi>K</mi></mfrac><mrow data-mjx-texclass="CLOSE"><mo minsize="1.2em" maxsize="1.2em">)</mo></mrow></mrow></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">\(\binom{N}{K}\)</span></mjx-container>를 10,007로 나눈 나머지를 출력한다.</p>

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import sys
input = sys.stdin.readline
N, K = map(int, input().split())
D = [[0 for j in range(N+1)] for i in range(N+1)]
for i in range(0, N+1):
D[i][0] = 1
D[i][i] = 1
D[i][1] = i
for i in range(1, N+1):
for j in range(1, i+1):
D[i][j] = (D[i-1][j-1] % 10007 + D[i-1][j] %10007)%10007
print(D[N][K])