poj 2100(尺取法)

Graveyard Design

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 6107		Accepted: 1444
Case Time Limit: 2000MS

Description

King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.
After a consultation with his astrologer, King George decided that
the lengths of section sides must be a sequence of successive positive
integer numbers. A section with side length s contains s²
graves. George has estimated the total number of graves that will be
located on the graveyard and now wants to know all possible graveyard
designs satisfying the condition. You were asked to find them.

Input

Input file contains n --- the number of graves to be located in the graveyard (1 <= n <= 10¹⁴ ).

Output

On
the first line of the output file print k --- the number of possible
graveyard designs. Next k lines must contain the descriptions of the
graveyards. Each line must start with l --- the number of sections in
the corresponding graveyard, followed by l integers --- the lengths of
section sides (successive positive integer numbers). Output line's in
descending order of l.

Sample Input

2030

Sample Output

2
4 21 22 23 24
3 25 26 27

题意:给你一个数，询问有多少种连续自然数的平方和等于这个数，输出所有可能
题解:尺取法遍历所有符合条件的区间，满足的话记录左边界以及右边界，计数器+1。
尺取法过程:

　　整个过程分为4布：

　　　　1.初始化左右端点

　　　　2.不断扩大右端点，直到满足条件

　　　　3.如果第二步中无法满足条件，则终止，否则更新结果

　　　　4.将左端点扩大1，然后回到第二步

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
using namespace std;
typedef long long LL;
LL n;
struct Node{
    LL l,r;
}res[];
int main()
{
    while(scanf("%lld",&n)!=EOF){

        LL l=,r=;
        LL len = (int)sqrt(n*1.0)+;
        LL sum = ;
        int cnt=;
        while(l<=len){
            while(r<=len&&sum<n){
                sum+=r*r;
                r++;
            }
            if(sum<n) break;
            if(sum==n){
                cnt++;
                res[cnt].l = l;
                res[cnt].r = r;
            }
            sum-=l*l;
            l++;
        }
        printf("%d\n",cnt);
        for(int i=;i<=cnt;i++){
            printf("%d ",res[i].r-res[i].l);
            for(int j=res[i].l;j<res[i].r-;j++){
                printf("%d ",j);
            }
            printf("%d\n",res[i].r-);
        }
    }
    return ;
}