結果
問題 | No.2603 Tone Correction |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 33 |
ソースコード
#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'; }