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CodeForces - 606A
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 colorinto 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 ab 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, xy 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

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.


初做题目的感想:【算法】算法是一种思维

英语理解问题
题目要求没读懂 以为是问上面的数据能否通过变换变成下面的数据
结果是只需要大于等于下面的就好了 -.-
所以我说我怎么一直 WA 
英语是硬伤啊 
然后读懂之后……
这题好水啊……

我说怎么一个个AC这么快


#include<stdio.h>
int main() {
	int a,b,c,x,y,z;
	while(scanf("%d %d %d",&a,&b,&c)!=EOF) {
		scanf("%d %d %d",&x,&y,&z) ;
		int sum1=0;
		int sum2=0;
		if(a>x)
			sum1+=(a-x)/2;
		else
			sum2+=x-a;
		if(b>y)
			sum1+=(b-y)/2;
		else
			sum2+=y-b;
		if(c>z)
			sum1+=(c-z)/2;
		else
			sum2+=z-c;
		if(sum1<sum2)
			printf("No\n");
		else
			printf("Yes\n");
	}
	return 0;
}

 所以决定恶补英语-.-


题目地址:【Codeforces】[606A]Magic Spheres