POJ：2100-Graveyard Design（尺取）

Graveyard Design

Time Limit: 10000MS Memory Limit: 64000K

Total Submissions: 8504 Accepted: 2126

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

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解题心得：

1. 给你一个数，问你他可以由多少个连续数的平方和得到，输出第一行为方案数，然后每一行第一个数是由多少个连续数的平方和得到，然后是连续数。
2. 直接跑尺取，满足尺取的两个特点。

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#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <vector>
using namespace std;
typedef long long ll;
vector < pair<ll,ll> > ve;
void solve(ll n) {
    ll l = 1,r = 1,sum = 0;

    while(1) {
        while(sum < n && r <= 1e7) {
            sum += r*r;
            r++;
        }
        if(sum < n)
            break;
        if(sum == n)
            ve.push_back(make_pair(l,r-1));
        sum -= l*l;
        l++;
        if(l*l > n)
            break;
    }
    printf("%lld\n",ve.size());
    for(ll i=0;i<ve.size();i++) {
        printf("%d ",ve[i].second-ve[i].first+1);
        for(ll j=ve[i].first;j<=ve[i].second;j++) {
            if(j == ve[i].first)
                printf("%lld",j);
            else
                printf(" %lld",j);
        }
        printf("\n");
    }
}

int main() {
    ll n;
    scanf("%lld",&n);
    solve(n);
    return 0;
}