Christmas period, the increase in the demand of orbs minmin senior would like to take this opportunity to cash, it is 9102 now, we can easily use for synthesis of magic magic crystal ball,

Min min senior want to get some yellow, green and blue orbs, known to synthesize a yellow orbs need two yellow crystal, synthesis of a green orbs need a yellow crystals and a blue crystal, synthesis of a blue orbs need three blue crystal

Now min min senior has A yellow crystal and B blue crystal, min min senior want A yellow orbs x, y, A green orbs and z A blue orbs, min min senior, please calculate the minimum also need the number of crystal, the first line contains two Numbers A and B (0? Or less? A,? B? Or less? 10 ^ 9), representing the existing number of yellow crystal, and the number of blue crystal,

The second line has three Numbers x, y, and z (0? Or less? X,? Y,? Z? Or less? 10 ^ 9) - represents the final number of yellow, green and blue magic ball to get
Input
4 3
2 1 1
The Output
2
Input
3 and 9
1 1 3
The Output
1
Input
12345678, 87654321,
43043751 1000000000 53798715
The Output
2147483648



I have made the wrong code,
#include
Int main ()
{unsigned long long int a, b, x, y, z, n, m, blue=0, yellow=0, t;
The scanf (" % LLD % LLD, & amp; A, & amp; B);

The scanf (" % LLD % % LLD LLD ", & amp; X, & amp; Y, & amp; Z);

N=b/3;
T=b % 3.
If (n<=z)
Blue=3 * (z - n) + y - t;
If (n> Z& & (n - z) & lt;=y)
Blue=y - (n - z) - t;
If (n> Z& & (n - z) & gt; Y)
Blue=0;
If (a<=y)
Yellow=y - a + 2 * x;
If (a> Y& & (a - y) % 2==0)
Yellow=(x - (a - y)/2) * 2;
If (a> Y& & (a - y) % 2!=0)
Yellow=(x - (a - y)/2) * 2-1;
M=blue + yellow;
Printf (" % LLD ", m);
return 0;
}

CodePudding user response:

Your program defines a heap of integer,
N=b/3; Are you sure that there will be no problem?
You how so trouble don't want to complex the problem of simple
For your reference

 int main () {
int a,b; 
While (the scanf (" % d % d ", & amp; A, & amp; B)!=(EOF) 
{
Unsigned int x, y, z; 
The scanf (" % u % u % du ", & amp; X, & amp; Y, & amp; Z); 
Printf (" % u \ n ", 2 * (x + y) + 3 * z - a - b). 
} 
}