链接:
https://codeforces.com/contest/1221/problem/A
题意:
You are playing a variation of game 2048. Initially you have a multiset s of n integers. Every integer in this multiset is a power of two.
You may perform any number (possibly, zero) operations with this multiset.
During each operation you choose two equal integers from s, remove them from s and insert the number equal to their sum into s.
For example, if s={1,2,1,1,4,2,2} and you choose integers 2 and 2, then the multiset becomes {1,1,1,4,4,2}.
You win if the number 2048 belongs to your multiset. For example, if s={1024,512,512,4} you can win as follows: choose 512 and 512, your multiset turns into {1024,1024,4}. Then choose 1024 and 1024, your multiset turns into {2048,4} and you win.
You have to determine if you can win this game.
You have to answer q independent queries.
思路:
将不大于2048的加起来判断
代码:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t, a, b, c;
cin >> t;
while (t--)
{
cin >> a >> b >> c;
if (a < b)
swap(a, b);
c += (a-b);
a = b;
int sum = min(a, c);
a -= sum, b -= sum, c -= sum;
cout << sum+(a+b)/3 << endl;
}
return 0;
}