USO

Ultra Spiritual

Being ultra spiritual has nothing to do with actually being spiritual.

Being ultra spiritual means you look spiritual

First Blood

'

C
or
Python?

Python

print('\n'.join(list(map(lambda k : ["Yes, I can.","USO!","No, I can't."][int(int(k[0])
*2//3+1<=int(k[1]))+int(int(k[0])*2//3+1==int(k[1]))],[input().split() for _ in range(int
(input()))]))))

C

#include<stdio.h>
int main(){
    int T,N,S;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&S);
        if(N*2/3+1 == S) printf("No, I can't.\n");
        else if(N*2/3 >= S) printf("Yes, I can.\n");
        else printf("USO!\n");
    }
}

Spiorad Algorithm

with your ultra spiritual

Step I -- lower bound

I accidentally make it happen.

N=3n

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00

N=3n

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00
0
1
n-1
n
n+1
n+2
2n-1
2n

N=3n

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00
0 n
1 n+1
n-1 2n-1
n 2n
n+1 0
n+2 1
2n-1 n-2
2n n-1

N=3n

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00
0 n n
1 n+1 n+2
n-1 2n-1 3n-2
n 2n 3n
n+1 0 n+1
n+2 1 n+3
2n-1 n-2 3n-3
2n n-1 3n-1
Ans:{2 \over 3}N + 1

N=3n+1

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00
0 n n
1 n+1 n+2
n-1 2n-1 3n-2
n 2n 3n
n+1 0 n+1
n+2 1 n+3
2n-1 n-2 3n-3
2n n-1 3n-1
Ans: \lfloor{2 \over 3} N \rfloor+ 1

N=3n+2

9:00~12:00 13:00~16:00 col1 + col2 17:00~20:00
0 n n
1 n+1 n+2
n 2n 3n
n+1 2n+1 3n+2
n+2 1 n+3
n+3 2 n+5
2n n-1 3n-1
2n+1 n 3n+1
Ans: \lfloor{2 \over 3} N \rfloor+ 1

Step II -- upper bound

Not exactly important

Assume we can tolerate at most k gamers.

9:00~12:00 13:00~16:00 17:00~20:00
a1 b1 c1
a2 b2 c2
ak bk ck

Assume we can tolerate at most k gamers.

9:00~12:00 13:00~16:00 17:00~20:00
a1 b1 c1
a2 b2 c2
ak bk ck
  • For one column, each number can't appear twice.
  • Hint for RHS : 
\sum\limits_{i=1}^{k}a_i \ge {k(k-1)\over2}
Assume\ a_i\in \{x\ |\ 0\leq x\lt k\}

Assume we can tolerate at most k gamers.

9:00~12:00 13:00~16:00 17:00~20:00
a1 b1 c1
a2 b2 c2
ak bk ck
\sum\limits_{i=1}^{k}a_i+\sum\limits_{i=1}^{k}b_i+\sum\limits_{i=1}^{k}c_i \ge 3\times {k(k-1)\over2}

Assume we can tolerate at most k gamers.

\sum\limits_{i=1}^{k}a_i+\sum\limits_{i=1}^{k}b_i+\sum\limits_{i=1}^{k}c_i \ge 3\times {k(k-1)\over2}
kN \ge 3\times {k(k-1)\over2}
k \le \lfloor {2\over 3}N\rfloor+1
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