Codeforces Round #456 Div.2 A. Tricky Alchemy

Problem

http://codeforces.com/contest/912/problem/A

Analysis

One needs 2·x + y yellow and 3·z + y blue crystals. The answer therefore is max(0, 2·x + y - A) + max(0, 3·z + y - B).

Code

C++

#include 
using namespace std;

int main() 
{
    long long A, B;
    long long x, y, z;
    cin >> A >> B;
    cin >> x >> y >> z; 
    
    long long yellow = 2 * x + y - A;
    long long blue = y + 3 * z - B, 0LL;
    
    cout << max(yellow, 0) + max(blue, 0);
    
    return 0;   
}

Python2

a, b = map(int, raw_input().split())
x, y, z = map(int, raw_input().split())

yellow = 2 * x + y
blue = y + 3 * z
ans = max(0, yellow - a) + max(0, blue - b)

print ans

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