牛客练习赛22 简单瞎搞题

简单瞎搞题

思路：

bitset写dp

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e6 + 5;
bitset<N> dp[2];
int main() {
    int n, l, r;
    scanf("%d", &n);
    int now = 0;
    dp[now].set(0);
    for (int i = 0; i < n; i++) {
        scanf("%d %d", &l, &r);
        now ^= 1;
        dp[now].reset();
        for (int j = l; j <= r; j++) {
            dp[now] |= dp[now^1] << (j*j);
        }
    }
    printf("%d
", dp[now].count());
    return 0;
}