loj515 「LibreOJ β Round #2」贪心只能过样例

传送门：https://loj.ac/problem/515

【题解】

容易发现S最大到1000000。

于是我们有一个$O(n^2*S)$的dp做法。

容易发现可以被bitset优化。

于是复杂度就是$O(\frac{n^2S}{32})$

然后……就过了

# include <bitset>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 1e6 + 10, N = 100 + 5;
const int mod = 1e9+7;

int n, a[N], b[N], cnt;

bitset<1000001> f[2];

int main() {
    cin >> n;
    for (int i=1; i<=n; ++i) cin >> a[i] >> b[i];
    int pre = 0, cur = 1;
    f[0][0] = 1;
    for (int i=1; i<=n; ++i) {
        f[cur].reset();
        for (int j=a[i]; j<=b[i]; ++j) 
            f[cur] |= (f[pre] << j*j);
        swap(cur, pre);
    }
    for (int i=0; i<=1000000; ++i)
        if(f[pre][i]) ++cnt;
    cout << cnt << endl;
    return 0;
}