2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 E. Half-consecutive Numbers

The numbers 11, 33, 66, 1010, 1515, 2121, 2828, 3636, 4545 and t_i=\frac{1}{2}i(i+1)ti=21i(i+1), are called half-consecutive.

For given NN, find the smallest rr which is no smaller than NN such that t_rtr is square.

Input Format

The input contains multiple test cases.

The first line of a multiple input is an integer TT followed by TT input lines.

Each line contains an integer N~(1\le N\le 10^{16})N (1≤N≤1016).

Output Format

For each test case, output the case number first.

Then for given NN, output the smallest rr.

If this half-consecutive number does not exist, output -1−1.

样例输入

样例输出

Case #1: 1
Case #2: 8
Case #3: 49
Case #4: 288

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
/*int main()
{
    long long i,t,tt,l;
    for(i=1;i<=10e16;i++)
    {
        t=i*(i+1);
        tt=t/2;
        l=sqrt(tt);//判断是否可以开平方!!!!!!!!
        if(l*l==tt)
            cout<<i<<endl;
        else
            continue;
    }
}直接枚举会超时12数个之后很久没出第十三个*/
/*int main()
{
   long long i,t,tt,l;
   for(i=1;i<=10e16;i++)
   {
       t=i*(i+1);
        tt=t/2;
        l=sqrt(tt);//判断是否可以开平方!!!!!!!!
        if(l*l==tt)
            cout<<i<<" "<<tt<<" "<<l<<endl;
        else
            continue;
   }
}l有规律，l i之间也有规律。。。。*/
/*int main()
{
    long long a[30],b[30];
    a[1]=1;
    a[2]=6;
    for(int i=3;i<=25;i++)
        a[i]=a[i-1]*6-a[i-2];
    b[1]=1;b[2]=8;b[3]=49;b[4]=288;
    long long k=14;
    for(int i=4;i<=25;i++)
    {
        k+=a[i-1]*2;
        b[i]=a[i]+k;
    }
}*/
//三角形数第n个三角形数sn=n*(n+1)/2
//既是三角形数又是完全平方数（正方形数）an=6*a[i-1]-a[i-2]+2
#define ll long long
int main()
{
    ll n;
    int t,cnt;
    ll a[25]={1,8,49,288,1681,9800,57121,332928,1940449,11309768,65918161,384199200,2239277041,13051463048,76069501249,443365544448,2584123765441,15061377048200,87784138523761,511643454094368,2982076586042449};
    cnt=1;
    cin>>t;
    while(t--)
    {
        scanf("%lld",&n);
        cout<<"Case #"<<cnt<<": ";
        if(n>2982076586042449)
        {
            cout<<-1<<endl;
            cnt++;
        }
        else
        {
            for(int i=0;i<=20;i++)
            {
              if(a[i]>=n)
              {
                  cout<<a[i]<<endl;
                  cnt++;
                  break;
              }
            }
        }
    }
}