poj 2100 Graveyard Design

/*
* @Author: Lyucheng
* @Date:   2017-08-02 16:43:37
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-08-02 21:22:15
*/
/*
 题意：给你一个10^14的范围内的n,让你判断有多少连续子序列的平方和是n

 思路：尺取
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define LL long long
#define INF 0xffffffff
#define MAXN 10000001

using namespace std;

LL n;
LL l,r;
LL _min;
LL pos;

struct  Node{
    int l,r;
    Node(){}
    Node(int _l,int _r):l(_l),r(_r){}
};
vector<Node>v;

void init(){
    v.clear();
    pos=0;
}

int main(){ 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%I64d",&n);
    int cnt=((int)sqrt(n*1.0)) +1;
    init();

    l=1,r=1;
    _min=INF;
    while(r*r<=n&&l*l<=n){
        while(pos<n&&r*r<=n){
            pos+=r*r;
            r++; 
        }
        while(pos>n){
            pos-=l*l;
            l++;
        }
        if(pos==n){
            v.push_back(Node((int)l,(int)r-1));
            pos-=l*l;
            l++;  
        }
    }
    int tol=v.size();
    printf("%d\n",tol);
    for(int i=0;i<tol;i++){
        printf("%d ",v[i].r-v[i].l+1);
        for(int j=v[i].l;j<=v[i].r;j++){
            printf(j==v[i].l?"%d":" %d",j);
        }
        printf("\n");
    }
    return 0;
}