POJ2100 Graveyard Design【尺取法】

Graveyard Design

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 7094		Accepted: 1733
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

Source

Northeastern Europe 2004, Northern Subregion

问题链接：POJ2100 Graveyard Design。

题意简述：对于输入的n，求一段连续的正整数，使得其平方和等于n。

问题分析：

采用尺取法解决。

step1.从1开始，先求出最前面的子序列，使之平方和大于或等于n；

step2.重复step3和step4，直到整数的平方小于n为止；

step3.若子序列平方和等于n，则先把解放入向量变量中；去掉子序列的第1个元素，子序列的后面再加上后续整数的平方和；

step4.将向量中的解取出，输出结果。

程序说明：需要注意数据类型，需要注意起始整数。

AC的C++语言程序如下：

/* POJ2100 Graveyard Design */

#include <iostream>
#include <vector>
#include <stdio.h>

using namespace std;

vector<pair<long long, long long> > ans;

void solve(long long n)
{
    ans.clear();

    long long start = 1, end = 1, sum = 0;
    while (start * start <= n) {
        while (end * end <= n && sum < n) {
            sum += end * end;
            end++;
        }

        if (sum == n)
            ans.push_back(make_pair(start, end));

        sum -= start * start;
        start++;
    }

    int len = ans.size();
    printf("%d
", len);
    for (int i = 0; i < len; i++) {
        printf("%lld", ans[i].second - ans[i].first);
        for (int j = ans[i].first; j < ans[i].second; j++)
            printf(" %d", j);
        printf("
");
    }
}

int main()
{
    long long n;

    while(scanf("%lld", &n) != EOF)
        solve(n);

    return 0;
}