Codeforces Round #753 Banquet Preparations 1 Solution (Java/C++)

Solution:

We represent the total remain weight of each dish by $remain_i$, which obviously has $remain_i=a_i+b_i-m$.

So in order to balance a and b as much as possible, obviously our goal is $\sum a_i = \frac {\sum remain_i} 2$.
At the same time, we can notice that the range of the remaining a must be in $[\min(remain_i, a_i), reman_i-\min(remain_i, b_i)]$.
Therefore, we can get the maximum and minimum values of $\sum a_i$. We only need to adjust the number of remained a for each dish according to the difference between $\sum a_i$ and $\frac {\sum remain_i} 2$.

Code:

Java

Submission #134367469 - Codeforces
Codeforces. Programming competitions and contests, programming community

C++

Submission #134367675 - Codeforces
Codeforces. Programming competitions and contests, programming community
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