POJ - 2100 - Graveyard Design(尺取)

题目链接：https://vjudge.net/problem/POJ-2100#author=DlPF

题目大意：给出一个数n。你要求一段连续的数，这些数的平方和等于n。

直接尺取，因为n最大1e14，所以尺取到1e7肯定出结果

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<stdio.h>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define mst(a) memset(a, 0, sizeof(a))
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const double pi = acos(-1.0);
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const int _NAN = -0x3f3f3f3f;
const int maxn = 1e7+10;
int* ans = new int[maxn];
int head[maxn/10];
int main(void) {
    ll num;
    scanf("%lld", &num);
    ll l = 1, r = 1, sum = 0, times = INF, cnt = 0, kase = 0;
    while(times--) {
        while(sum < num && r*r <= num) {
            sum += r*r;
            ++r;
        }
        if (sum < num)
            break;
        if (sum == num) {
            head[cnt++] = r-l;
            for (int i = l; i<r; ++i)
                ans[kase++] = i;
        }
        sum -= l*l;
        sum = max(sum, 0);
        ++l;
    }
    int len = 0;
    printf("%lld\n", cnt);
    for (int i = 0; i<cnt; ++i) {
        printf("%d ", head[i]);
        len += head[i];
        for (int j = len-head[i]; j<len; ++j)
            printf(j == len-1 ? "%d\n" : "%d ", ans[j]);
    }
    return 0;
}