| Time Limit: 2000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
Carl is a beginner magician. He has a blue,
b violet and c orange magic spheres. In one move he can transform two spheres
of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least
x blue, y violet and
z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a,
b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x,
y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Sample Input
4 4 0
2 1 2
Yes
5 6 1
2 7 2
No
3 3 3
2 2 2
Yes
Sample Output
Hint
In the first sample the wizard has 4 blue and
4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have
2 blue and 5 violet spheres. Then he turns
4 violet spheres into 2 orange spheres and he ends up with
2 blue, 1 violet and
2 orange spheres, which is exactly what he needs.
Source
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a,b,c,x,y,z;
long long ans,pre;
int main()
{
while(scanf("%d%d%d%d%d%d",&a,&b,&c,&x,&y,&z)!=EOF)
{
ans=pre=0;
int p,q,r;
p=q=r=0;
if(a-x>=0) p=0,pre+=(a-x)/2;
else p=a-x;
if(b-y>=0) q=0,pre+=(b-y)/2;
else q=b-y;
if(c-z>=0) r=0,pre+=(c-z)/2;
else r=c-z;
// printf("%d\n",pre);
// printf("%d %d %d\n",p,q,r);
if(pre+p+q+r>=0) printf("Yes\n");
else printf("No\n");
}
return 0;
}