poj 2100 尺取法 一个数字拆成连续数字平方和

题意：将一个数拆成若干个连续数字的平方和。

用尺取法枚举区间，复杂度为O(n)，时限10s，3s多ac。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const int N = 100;

struct Node
{
    int from, to;
} node[N];

void solve( long long n )
{
    int u = sqrt( n * 1.0 ), cnt = 0, q = 1;
    long long sum = 0;
    for ( int i = 1; i <= u; i++ )
    {
        sum += ( long long ) i * i;
        while ( sum > n )
        {
            sum -= ( long long ) q * q;
            q++;
        }
        if ( sum == n )
        {
            node[cnt].from = q;
            node[cnt].to = i;
            cnt++;
        }
    }
    printf("%d
", cnt);
    for ( int i = 0; i < cnt; i++ )
    {
        printf("%d", node[i].to - node[i].from + 1);
        for ( int j = node[i].from; j <= node[i].to; j++ )
        {
            printf(" %d", j);
        }
        puts("");
    }
}

int main ()
{
    long long n;
    while ( scanf("%lld", &n) != EOF )
    {
        solve(n);
    }
    return 0;
}