#364524 (C) No.270 next_permutation (1)

• 問題: No.270 next_permutation (1)
• ユーザー: bal4u
• 提出日時: 2019-08-01 11:11:38
• 言語: C (gcc 13.3.0)
• 結果: AC
• 実行時間: 7 ms / 2,000 ms
• コード長: 1,743 bytes
• コンパイル時間: 237 ms
• コンパイル使用メモリ: 30,720 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-07-05 07:27:00
• 合計ジャッジ時間: 1,157 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	AC * 15

コンパイルメッセージ

main.c: In function 'in':
main.c:9:14: warning: implicit declaration of function 'getchar_unlocked' [-Wimplicit-function-declaration]
    9 | #define gc() getchar_unlocked()
      |              ^~~~~~~~~~~~~~~~
main.c:14:24: note: in expansion of macro 'gc'
   14 |         int n = 0, c = gc();
      |                        ^~

ソースコード

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// yukicoder: No.270 next_permutation (1)
// 2019.8.1 bal4u
#include <stdio.h>
typedef long long ll;
#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif
int in() {   // 非負整数の入力
    int n = 0, c = gc();
    //  while (isspace(c)) c = gc();
    do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
    return n;
}
#define ABS(x)  ((x)>=0?(x):-(x))
int p[100005]; int N;
int b[100005];
ll sum;
void swap(int i, int j) {
    int t;
    sum -= ABS(b[i]-p[i]) + ABS(b[j]-p[j]);
    t = p[i], p[i] = p[j], p[j] = t;
    sum += ABS(b[i]-p[i]) + ABS(b[j]-p[j]);
}
/* algorithm of next_permutation in lexicographic order
    Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
    Find the largest index l greater than k such that a[k] < a[l].
    Swap the value of a[k] with that of a[l].
    Reverse the sequence from a[k + 1] up to and including the final element a[n].
*/
void next_permutation(void) {
    int k, l, r;
    // 右端にある最も長い降順部分列 p[k+1]... p[N-1]をみつける
    k = N-2; while (k >= 0 && p[k] > p[k+1]) k--;
    
    // 上記の降順部分列を昇順列に戻す
    l = k+1, r = N-1;
    while (l < r) swap(l, r), l++, r--;
    
    if (k < 0) return;
    
    //p[k]より大きい要素 p[r]を見つけ、交換する
    r = k+1; while (p[k] > p[r]) r++;
    swap(r, k);
}
int main()
{
    int i, K, f;
    ll ans;
    N = in(), K = in();
    if (K == 0) { puts("0"); return 0; }
    for (i = 0; i < N; i++) p[i] = in();
    sum = 0, f = 0;
    for (i = 0; i < N; i++) {
        b[i] = in(), sum += ABS(b[i]-p[i]);
        if (b[i] < N) f = 1;
    }
    if (f) {
        ans = 0;
        while (K--) ans += sum, next_permutation();
        printf("%lld\n", ans);
    } else printf("%lld\n", sum*K);
    return 0;
}
 
 
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