NUC1840 Graveyard Design【尺取法】

Graveyard Design

时间限制: 10000ms 内存限制: 64000KB

通过次数: 1总提交次数: 1

问题描述

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.

输入描述

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

输出描述

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.

样例输入

2030

样例输出

2
4 21 22 23 24
3 25 26 27

来源

{Northeastern Europe 2004}, Northern Subregion

问题分析：（略）

这个问题和《POJ2100 Graveyard Design【尺取法】》是同一个问题，代码直接用就AC了。

程序说明：参见参考链接。

参考链接：POJ2100 Graveyard Design【尺取法】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++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\n", 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("\n");
    }
}

int main()
{
    long long n;

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

    return 0;
}