poj 2100 Graveyard Design

                                                                                                                                                          Graveyard Design

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 6597		Accepted: 1588
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:国王要建造墓地，墓地可以分n块区域(n=1,2...),且区域是线性排列的，区域从前到后依次分配连续自然数的平方块墓碑。现在给你需要放置的墓碑的总数，让你给出所有可能的分配情况。
思路：尺取法。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef unsigned long long ll;
const int N_MAX = 100000;
bool what;
ll num;
vector<pair<ll, ll>>answer;
int main() {
    ll n;
    while (scanf("%lld", &n) != EOF) {
        num = 0;
        ll l = 1, r = 1, sum = 0;
        while (1) {
            for (;;) {
                while ( r*r<=n&&sum < n) {
                    sum += r*r;
                    r++;
                }
                if (sum < n) {
                    what = 0;
                    break;
                }
                if (sum == n) {
                    /////记录数据
                    answer.push_back(make_pair(l,r-1));
                    /////
                    what = 1;
                    num++;
                    break;
                }
                sum -= l*l;
                l++;
            }
            if (what) {
                l++;
                r = l;
                sum = 0;
            }
            else break;
        }
        ///////
        printf("%lld
",num);
        for (vector<pair<ll, ll>>::iterator it = answer.begin(); it != answer.end(); it++) {
            printf("%lld",it->second-it->first+1);
            for (ll j =it->first; j <= it->second; j++)
                printf("% lld", j);
            printf("
");
        }
        ///////////
    }
    return 0;
}