@yjkim997

#include <stdio.h> int n, a[200010], b[200010]; int main() { scanf("%d", &n); for(int i=0; i<n; i++) scanf("%d", &a[i]); for(int j=0; j<n; j++) scanf("%d", &b[j]); int aidx=0, bidx=0, aemp=-1, bemp=-1; while(aidx<n||bidx<n) {

/* //CODEUP 2413 #include <stdio.h> int K, HH, MM, time=0; int main() { scanf("%d", &K); scanf("%d %d", &HH, &MM); if(K<1200) printf("%2d:%2d +%d",HH,MM,0);

# Prüfer Code Decoder # Decoder 소개 print("[길이 (n-2)의 Prüfer Code를 입력받아, 원래 수형도 T를 복구하는 프로그램입니다.]\n") while True : try : n=int(input("수형도 T의 꼭짓점 개수(n)를 입력하세요 : ")) if n < 2 : print("유효하지 않은 입력입니다. n은 2 이상이어야 합니다.\n")

#include <stdio.h> #include <vector> #define MAX_N 10000 #define MAX_M 1000 using namespace std; int N, M, adj[MAX_N][MAX_M]; vector<int> mMatch, wMatch; vector<bool> visited;

# Hall's Marriage Theorem Simulation ''' 참고자료 [1] https://gazelle-and-cs.tistory.com/10 [2] 알고리즘 문제해결 전략 2 ''' n=int(input("전체 남성의 수를 입력하세요.(예. 1, 2, 3...) : ")) print("i번째 남성이 알고 있는 여성의 번호를 i번째 줄에 공백으로 구분지어 입력하세요.") print("(예. 1번 남성이 알고 있는 여성의 번호가

#include <stdio.h> #include <stdlib.h> #include <string.h> struct Data { char name[25]; } struct Tnode { Tnode* left=NULL, right=NULL; Data d; } struct BST { private : Tnode *root=NULL; public : Tnode* getRoot() { return root; }

#include <stdio.h> #include <algorithm> #include <stack> #include <vector> using namespace std; int n, arr[300010], INDEX[300010]; vector<int> v; stack<int> st; int main() {

# 292 s="aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

//-------------------------------------------------------------------------------- //SASAOJ-A_[함수+포인터] swap 함수 만들기 #include <stdio.h> // void myswap(int *a, int *b) { if(*a<*b) return; int tmp=*a; *a=*b; *b=tmp;

#include <stdio.h> int arr[25][25], row[25], col[25]; struct Pos{ int r, c; }; int f(Pos s, Pos e, bool rc){ int cnt = 0; bool c = 1; for(int i=s.r; i<=e.r; i++){

#include <iostream> using namespace std; double max(double a, double b) { return a>b?a:b; } int n, Ly[2], Lm[2], Ld[2], LD[2], mon[]={0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int y[110], m[110], d[110], res=0; int main() { ios_b

#include <stdio.h> int prime[1000010]={1, 1, }; void Prime() { for(int i=2; i<=1000; i++) { if(prime[i]==0) { for(int j=i*i; j<=1000000; j+=i) prime[j]=1; } }

#include <stdio.h> #include <vector> #include <algorithm> #define ll long long struct candy { ll amount; int cnt; };

#include <stdio.h> #include <vector> #include <algorithm> #define ll long long #define MOD 998244353 struct Edge { int from, to; int idx; };

#include <stdio.h> #include <vector> int p[2010][2010], pos[2010], res; int main() { int n, m; scanf("%d %d", &n, &m); for(int i=0; i<n; i++) { for(int j=0; j<m; j++) {

#include <stdio.h> #include <vector> #include <queue> int indegree[1010]; int G[1010][1010]; int main() { int n, m, vcnt=0; scanf("%d %d", &n, &m);

/* // n이 홀수일 때 로직 다시 짜기 #include <stdio.h> int main() { int t; scanf("%d", &t); while(t--) { int n;

#include <stdio.h> #include <deque> int arr[50], ptr=0; int main() { long long n, k; scanf("%lld %lld", &n, &k); std::deque<char> dq;

//슬프다 엉엉.. O(1)으로 탐색할 방법을 너무 늦게 알았다.. //TLE 날 건 알고 있었는데 T.T #include <stdio.h> #include <map> #include <vector> #define ll long long using namespace std; struct block { int idx;

#include <stdio.h> #include <vector> using namespace std; int main() { int t; scanf("%d", &t); while(t--) { int n, p[200010]={}, q[200010]={};