結果
| 問題 | 
                            No.197 手品
                             | 
                    
| コンテスト | |
| ユーザー | 
                             sue_charo
                         | 
                    
| 提出日時 | 2017-04-17 18:19:12 | 
| 言語 | Python3  (3.13.1 + numpy 2.2.1 + scipy 1.14.1)  | 
                    
| 結果 | 
                             
                                AC
                                 
                             
                            
                         | 
                    
| 実行時間 | 38 ms / 1,000 ms | 
| コード長 | 1,718 bytes | 
| コンパイル時間 | 111 ms | 
| コンパイル使用メモリ | 12,672 KB | 
| 実行使用メモリ | 11,520 KB | 
| 最終ジャッジ日時 | 2024-07-20 03:51:28 | 
| 合計ジャッジ時間 | 2,875 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge3 / judge5 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 4 | 
| other | AC * 43 | 
ソースコード
# coding: utf-8
import array, bisect, collections, heapq, itertools, math, random, re, string, sys, time
sys.setrecursionlimit(10 ** 7)
INF = 10 ** 20
MOD = 10 ** 9 + 7
 
 
def II(): return int(input())
def ILI(): return list(map(int, input().split()))
def IAI(LINE): return [ILI() for __ in range(LINE)]
def IDI(): return {key: value for key, value in ILI()}
 
 
def solve(S_bef, N, S_aft):
    bef_list = list(S_bef)
    aft_list = list(S_aft)
    bef_count = 0
    aft_count = 0
    for bef, aft in zip(bef_list, aft_list):
        if bef == "o":
            bef_count += 1
        if aft == "o":
            aft_count += 1
    
    if bef_count != aft_count:
        return "SUCCESS"
    else:
        if N == 0:
            if S_bef == S_aft:
                return "FAILURE"
            else:
                return "SUCCESS"
        elif N == 1:
            if bef_count == 1:
                if S_bef == "oxx" and S_aft == "xxo":
                    return "SUCCESS"
                elif S_bef == "xox" and S_aft == "xox":
                    return "SUCCESS"
                elif S_bef == "xxo" and S_aft == "oxx":
                    return "SUCCESS"
                
            elif bef_count == 2:
                if S_bef == "xoo" and S_aft == "oox":
                    return "SUCCESS"
                elif S_bef == "oxo" and S_aft == "oxo":
                    return "SUCCESS"
                elif S_bef == "oox" and S_aft == "xoo":
                    return "SUCCESS"
        
        else:
            return "FAILURE"
            
    return "FAILURE"
 
def main():
    S_bef = input()
    N = II()
    S_aft = input()
    print(solve(S_bef, N, S_aft))
 
 
if __name__ == "__main__":
    main()
            
            
            
        
            
sue_charo