POJ2100-Graveyard Design

Graveyard Design

Time Limit: 10000MS Memory Limit: 64000KTotal Submissions: 7036 Accepted: 1717Case 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

24 21 22 23 243 25 26 27

Source

Northeastern Europe 2004, Northern Subregion

题意：给你一个数，问有多少种一段连续的数的平方和等于这个数，并输出这些方案

解题思路：尺取法，当一段数的平方和等于这个数时，保存答案

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;LL ans[1005][100005];int main(){    LL n;    while(~scanf("%lld",&n))    {        int cnt=0;        LL head=0,rear=0,sum=0;        while(rear*rear<n)        {            rear++;            sum+=rear*rear;            while(sum>n)            {                head++;                sum-=head*head;            }            if(n==sum)            {                cnt++;                int k=1;                for(LL i=head+1;i<=rear;i++)                    ans[cnt][k++]=i;                ans[cnt][0]=k-1;            }        }        printf("%d\n",cnt);        for(int i=1;i<=cnt;i++)        {            printf("%d",ans[i][0]);            for(int j=1;j<=ans[i][0];j++)                printf(" %d",ans[i][j]);            printf("\n");        }    }    return 0;}