POJ - 2100 Graveyard Design

n虽然高达1e14，但是满足条件的s不会超过sqrt(n)。可以想到在O(sqrt(n))的复杂度下，

求一个[1,sqrt(n)]连续区间和为n的方案。

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

typedef long long ll;

const int N = 1e3;
int ans[N][2];

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    ll n;
    scanf("%I64d",&n);
    ll m = sqrt(n+0.5);
    ll sum = 0;
    int sz = 0;
    for(ll i = 1, j = 1; i <= m; i++){
        sum += i*i;
        while(sum > n) {
            sum -= j*j; j++;
        }
        if(sum == n) {
            ans[sz][0] = j;
            ans[sz++][1] = i;
        }
    }
    printf("%d\n", sz);
    for(int i = 0; i < sz; i++){
        printf("%d",ans[i][1]-ans[i][0]+1);
        for(int j = ans[i][0]; j <= ans[i][1]; j++){
            printf(" %d", j);
        }
        puts("");
    }
    return 0;
}