LibreOJ #515. 「LibreOJ β Round #2」贪心只能过样例

Posted 2020-10-01 ZlycerQan

二次联通门 : LibreOJ #515. 「LibreOJ β Round #2」贪心只能过样例

 

 

 

 

/*
    LibreOJ #515. 「LibreOJ β Round #2」贪心只能过样例
    
    很显然
    贪心方程哦不
    dp方程为 f[i][j]=f[i-1][j-k*k] 
    但是这样的话复杂度就是O(N ^ 5)
    
    那么就用bitset优化一下
    就ok了 
*/
#include <iostream>
#include <cstdio>
#include <bitset>

void read (int &now)
{
    register char word = getchar ();
    for (; !isdigit (word); word = getchar ());
    for (now = 0; isdigit (word); now = now * 10 + word - ‘0‘, word = getchar ());
}

#define Max 1000900

using namespace std;
bitset <Max> number[110];

int main (int argc, char *argv[])
{
    register int i, j;
    int N, l, r;
    
    read (N);
    number[0].set (0);
    for (i = 1; i <= N; ++ i)
    {
        number[i].reset ();
        read (l);
        read (r);
        for (j = l; j <= r; ++ j)
            number[i] |= (number[i - 1] << (j * j));
    }
    printf ("%d\n", number[N].count ());
    
    return 0;
}