codeforces1199C MP3（离散化前缀和二分）_随笔

codeforces1199C MP3（离散化/前缀和/二分）

题目链接：codeforces 1199C

题目思路：将 a [ 1 … n ] a[1…n] a[1…n] 离散化，前缀和维护区间个数。枚举区间起点，二分查找终点，取最大值。

参考代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int N = 4e5 + 10;
const int inf = 0x3f3f3f3f;
int n, I;
ll a[N], b[N], pre[N];
ll count(ll l ,ll r) {
    return pre[r] - pre[l-1];
}
bool check(ll l, ll r) {
    int k = r - l + 1;
    return ceil(log2(k)) * n <= I * 8;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n >> I;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(a+1, a+1+n);
    sort(b+1, b+1+n);
    ll m = unique(b+1, b+1+n) - b - 1;
    for (int i = 1; i <= n; i++) a[i] = lower_bound(b+1, b+1+m, a[i]) - b;
    for (int i = 1; i <= n; i++) {
        pre[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
        pre[i] += pre[i-1];
    }
    ll ans = inf;
    for (int i = 1; i <= m; i++) {
        ll l = i, r = m, mid;
        while (l < r) {
            mid = (l + r + 1) >> 1;
            if (check(i, mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }   
        }
        if (check(i, l)) {
            ans = min(ans, n - count(i, l));
        }
    }
    cout << ans << endl;
    return 0;
}

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codeforces1199C MP3（离散化前缀和二分）

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