Gym 100283F Bakkar In The Army

数学公式： n^2的前n项和n(n+1)(2*n+1)/6，用二分进行查找；

算出层数后继续二分查找位于这一层的哪一位，也可以推出相应公式

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

ll f(ll n)  //a(n)=n^2的前n项和公式
{
    return n*(n+1)*(2*n+1)/6;
}

ll f2(ll i,ll n)    //这个公式要推一下
{
    if(i<=n) return i*(i+1)/2;
    return n*n-(2*n-i)*(2*n-i-1)/2;
}

int main()
{
    freopen("army.in","r",stdin);
    int t;
    scanf("%d",&t);
    int ti=1;
    while(t--)
    {
        ll n;
        scanf("%I64d",&n);
        int lef=1,rig=1500000,floor;
        while(1)    //二分算哪一层
        {
            floor=(lef+rig)/2;
            if(n>f(floor-1) && n<=f(floor))
                break;
            if(n<=f(floor)) rig=floor-1;
            else lef=floor+1;
        }
        ll tmp=f(floor-1);
        lef=1;
        rig=2*floor-1;
        int index;
        while(1)    //二分算这一层的哪一个
        {
            index=(lef+rig)/2;
            if(n>tmp+f2(index-1,floor) && n<=tmp+f2(index,floor))
                break;
            if(n<=tmp+f2(index,floor)) rig=index-1;
            else lef=index+1;
        }
        printf("Case %d: %I64d
",ti++,(ll)(floor-1)*(floor-1)+index);
    }
    return 0;
}