Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2661

Solution Idea:

In this problem the number of warriors is given and we have to say how many row can be made with them in this rule that the row contains warriors equal to row number (i th row contains i warriors). so that if there is 3 row there we have 1+2+3 = 6 worriers.

We can use binary search to solve this problem. let low=0 and high= 10000000000 (A large value ) and binary search the ans which satisfy the formula n*(n+1)/2  according to given input.

/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll ask(ll n)
{
    return n*(n+1)/2;
}

int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    TEST_CASE(t)
    {
        ll x;
        cin>>x;
        ll lo=0,hi=10000000000,ans=0;

        while(lo<=hi)
        {
            ll mid=(lo+hi)/2;

            ll temp=ask(mid);

            if(temp<=x)
                {
                    ans=mid;
                    lo=mid+1;
                }
            else
                hi=mid-1;
        }

        cout<<ans<<endl;
    }
    return 0;
}