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rank 01: cheissmart | |
rank 02: FHVirus | |
rank 03: Ji_Kuai | |
rank 04: amano_hina | |
rank 05: alvingogo | |
rank 06: LittleCube | |
rank 07: 810848 | |
rank 08: nathanlee726 | |
rank 09: hhhhhaura | |
rank 10: Ejam |
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# MTF0000 題解 | |
## **unique1** | |
### 創作靈感 | |
在找工作時,接觸了一些公司為了徵才而準備的資料結構與演算法的面試題,發現面試題相對於一般的演算法競賽,更注重記憶體的使用量,除了優化時間複雜度外,也會希望面試者能優化記憶體使用量。所以覺得偶爾出一些在記憶體用量上找碴的題目對大家可能也有所幫助~ | |
同時,以測試一個新 Judge 平台的角度而言,我們也想準備一道能好好測試記憶體限制能否正常發揮的題目,於是此題就誕生啦! |
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--- | |
title: 出題是種態度 (駭客題) (Extreme) | |
time_limit: 3 | |
checker: checker | |
author: dreamoon | |
source: 經典問題 | |
... | |
# Description |
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#include<bits/stdc++.h> | |
using namespace std; | |
const int SIZE = 85; | |
char s[2][SIZE]; | |
int dp[SIZE][SIZE]; | |
int n[2]; | |
int nxt[2][SIZE][26]; | |
char an[SIZE]; | |
void dfs(int id1,int id2,int pos){ | |
if(dp[id1][id2]==0){ |
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#include <bits/stdc++.h> | |
#define SZ(X) ((int)(X).size()) | |
#define ALL(X) (X).begin(), (X).end() | |
#define REP(I, N) for (int I = 0; I < (N); ++I) | |
#define REPP(I, A, B) for (int I = (A); I < (B); ++I) | |
#define RI(X) scanf("%d", &(X)) | |
#define RII(X, Y) scanf("%d%d", &(X), &(Y)) | |
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z)) | |
#define DRI(X) int (X); scanf("%d", &X) | |
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y) |