poj 2100 Graveyard Design(尺取)

最新推荐文章于 2020-09-03 17:01:46 发布

cillyb

最新推荐文章于 2020-09-03 17:01:46 发布

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分类专栏：尺取文章标签：尺取算法

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本文介绍了一种用于设计由不同大小正方形区域组成的皇家墓地的算法。国王希望墓地包含多个不同规模的正方形墓区，且这些墓区边长为连续正整数。文章提供了完整的解决方案及代码实现。

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Graveyard Design

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 6385		Accepted: 1517
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 s ²graves. 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 <= 10 ¹⁴ ).

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

Sample Output

2
4 21 22 23 24
3 25 26 27

Source

Northeastern Europe 2004, Northern Subregion

要注意的是不能把平方和预处理出来，会MLE。

还有头和尾都要从1开始，不能从0开始。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxm = 1e5+5;
struct node
{
    long long l, r;
    node() {}
    node(long long ll, long long rr): l(ll), r(rr) {}
}res[maxm];

int main(void)
{
    long long n;
    while(cin >> n)
    {
        long long s = 1, e = 1, ans = 0;
        long long sum =  0;
        while(1)
        {
            while(e*e <= n && sum < n)
                sum += e*e, e++;
            if(sum < n) break;
            if(sum == n) res[ans++] = node(s, e);
            sum -= s*s, s++;
        }
        printf("%lld\n", ans);
        for(int i = 0; i < ans; i++)
        {
            printf("%d", res[i].r - res[i].l);
            for(int j = res[i].l; j < res[i].r; j++)
                printf(" %d", j);
            printf("\n");
        }
    }
    return 0;
}