F. Bakkar In The Army 二分

​​http://codeforces.com/gym/100283/problem/F​​

思路是二分第几行，二分出来的行是总和 >= n的，那么第k - 1行一定要选，那么再在第k行中二分那一列、

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
LL f1(LL n) {
    return n * (n + 1) / 2 * (2 * n + 1) / 3;
}
LL f2(int which, int pos) {
    if (pos <= which) {
        return 1LL * pos * (1 + pos) / 2;
    } else {
        LL t = 2 * which - 1 - pos;
        return 1LL * which * (1 + which) - which - f2(which, t);
    }
}
void work() {
    LL n;
    scanf("%I64d", &n);
    int be = 1, en = 2e6 + 20;
    while (be <= en) {
        int mid = (be + en) >> 1;
        LL res = f1(mid);
        if (res >= n) {
            en = mid - 1;
        } else be = mid + 1;
    }
//    cout << be << endl;
//    cout << f1(be) << endl;
    LL ans = ((LL)be - 1) * (be - 1);
//    cout << n << endl;
//    cout << f1(be - 1) << endl;
    n -= f1(be - 1);
//    cout << n << endl;
//    cout << f2(4, 8) << endl;
    int L = 1, R = 2 * be - 1;
    while (L <= R) {
        int mid = (L + R) >> 1;
        LL res = f2(be, mid);
        if (res >= n) {
            R = mid - 1;
        } else L = mid + 1;
    }
    ans += L;
    static int f = 0;
    printf("Case %d: %I64d\n", ++f, ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
//    freopen("army.in", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}

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