CodeForces--606A --Magic Spheres(模拟水题)

Magic Spheres

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

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Status

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

Input

4 4 0
2 1 2

Output

Yes

Input

5 6 1
2 7 2

Output

No

Input

3 3 3
2 2 2

Output

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

Codeforces Round #335 (Div. 2)

现在给你a,b,c个不同颜色的球，神魔颜色我也不知道，两个相同颜色的可以得到一个任意颜色的，问现在是否可以得到x,y,z,个三种颜色的球

#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;
}