E 简单瞎搞题

题目地址：

https://ac.nowcoder.com/acm/contest/5556/E

基本思路：

bitset优化dp，设记录的是到第个数时的所有可能情况的集合，那么比较容易得到转移方程： , 。

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

int n;
bitset<1000100> dp[110];
signed main() {
  IO;
  cin >> n;
  dp[0][0] = 1;
  rep(i,1,n){
    int l,r;
    cin >> l >> r;
    rep(j,l,r) dp[i] |= (dp[i-1] << (j * j));
  }
  cout << dp[n].count() << '\n';
  return 0;
}