solution

用表示前k个区间能(1)不能(0)组成这个数字。然后枚举一下第个区间所选的数字x，就有转移

最后数一下有多少数字可以被组成就行了。

但是这样复杂度爆炸，发现f的值只有和所以可以用进行优化，复杂度

code

/*
* @Author: wxyww
* @Date:   2020-05-19 14:37:41
* @Last Modified time: 2020-05-19 14:46:32
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
#include<bitset>
using namespace std;
typedef long long ll;
bitset<1000010>B,tmp,k;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}

int main() {
    int n = read();
    B[0] = 1;
    for(int i = 1;i <= n;++i) {
        int l = read(),r = read();
        tmp = B;
        B = k;
        for(int j = l;j <= r;++j) {
            B |= tmp << (j * j);
        }
    }
    int cnt = 0;
    for(int i = 0;i <= 1000000;++i) if(B[i]) cnt++;
    cout<<cnt;
    return 0;
}