結果

問題 No.2603 Tone Correction
ユーザー Jeroen Op de Beek
提出日時 2024-01-12 22:09:49
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 68 ms / 2,000 ms
コード長 1,581 bytes
コンパイル時間 2,265 ms
コンパイル使用メモリ 197,156 KB
最終ジャッジ日時 2025-02-18 18:22:31
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 33
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include "bits/stdc++.h"
using namespace std;
#define all(x) begin(x),end(x)
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream&
    operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#define debug(a) cerr << "(" << #a << ": " << a << ")\n";
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int mxN = 1e5+1, oo = 1e9;
/*
Take adjacent differences
Then want to do some greedy?
Can do +1 or -1 on one of them.
need to make everything 0
can balance out +1's and -1's.
can do +M or -M free of charge.
either needs +1's or -1's
can try all those options of switchings?
*/
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n,m; cin >> n>> m;
vi a(n),b(n);
for(auto& i : a) cin >> i;
for(auto& i : b) cin >> i;
vi c(n);
c[0]=(a[0]-b[0]+m)%m;
for(int i=1;i<n;++i) {
int cur = (a[i]-b[i]+m)%m;
int prv = (a[i-1]-b[i-1]+m)%m;
c[i] = (cur-prv+m)%m;
}
sort(all(c));
ll minus = accumulate(all(c),0LL);
ll plus=0;
ll ans = minus;
for(int i=n-1;i>=0;--i) {
auto cur = c[i];
if(cur) {
minus-=c[i];
plus+=m-c[i];
}
ans=min(ans,max(minus,plus));
}
cout << ans << '\n';
}
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