#463905 (Python3) No.39 桁の数字を入れ替え

• 問題: No.39 桁の数字を入れ替え
• ユーザー: tomoyaatcoder
• 提出日時: 2020-04-14 02:50:53
• 言語: Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1)
• 結果: RE
• 実行時間: -
• コード長: 3,595 bytes
• コンパイル時間: 152 ms
• コンパイル使用メモリ: 13,312 KB
• 実行使用メモリ: 11,904 KB
• 最終ジャッジ日時: 2024-10-01 09:34:39
• 合計ジャッジ時間: 1,516 ms
• ジャッジサーバーID （参考情報）: judge1 / judge5

ファイルパターン	結果
other	RE * 19

ソースコード

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import sys
from sys import stdin
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt
from functools import lru_cache, reduce
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10 ** 7)
# UnionFind
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        self.parents[x] += self.parents[y]
        self.parents[y] = x
    def size(self, x):
        return -self.parents[self.find(x)]
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
    def group_count(self):
        return len(self.roots())
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}
    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
# ダイクストラ
def dijkstra_heap(s, edge, n):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d
# 素因数分解
def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])
    if temp!=1:
        arr.append([temp, 1])
    if arr==[]:
        arr.append([n, 1])
    return arr
# 2数の最小公倍数
def lcm(x, y):
    return (x * y) // gcd(x, y)
# リストの要素の最小公倍数
def lcm_list(numbers):
    return reduce(lcm, numbers, 1)
# リストの要素の最大公約数
def gcd_list(numbers):
    return reduce(gcd, numbers)
# limit以下の素数を列挙
def eratosthenes(limit):
    A = [i for i in range(2, limit+1)]
    P = []
    
    while True:
        prime = min(A)
        
        if prime > sqrt(limit):
            break
            
        P.append(prime)
            
        i = 0
        while i < len(A):
            if A[i] % prime == 0:
                A.pop(i)
                continue
            i += 1
            
    for a in A:
        P.append(a)
            
    return P
# ここから書き始める
n = input()
nlen = len(n)
maximum = 1
maxcnt = 0
for i in range(1, nlen):
    if int(n[i]) > maximum:
        maximum = int(n[i])
        maxcnt = i
if maxcnt > 0:
    ans = str(maximum) + n[1:i] + n[0] + n[i + 1:]
    print(ans)
else:
    print(n)
 
 
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