CodeForces 606A Magic Spheres【水题】-优快云博客

本文链接：https://blog.youkuaiyun.com/Bcwan_/article/details/53042360

本文探讨了一道有趣的编程题目，魔术师需要通过变换不同颜色的球来完成特定任务。文章解析了解题思路，并提供了一个简洁的C++实现方案。

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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

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 2orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.

/*
    题意：有蓝色,紫色,橘色三种颜色球,你可以选择两个相同颜色的球将它们变成一个任意颜色的球.
          现在给出初始拥有的球数abc.问你能否得到球数xyz.
    类型：水题
    分析：因为可以选择两个相同颜色的球将它们变成一个任意颜色的球,所以当一种颜色的球数够了以后我们可以把多出来的
          数目num/2记录下来,然后判断是否总的多余能填补空缺
*/
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>
#include<cstring>
using namespace std;
typedef long long ll;
int main()
{
    int a,b,c,x,y,z,tmp=0;
    scanf("%d%d%d%d%d%d",&a,&b,&c,&x,&y,&z);
    if(a>x)tmp+=(a-x)/2;
    if(b>y)tmp+=(b-y)/2;
    if(c>z)tmp+=(c-z)/2;
    if(a<x)tmp-=x-a;
    if(b<y)tmp-=y-b;
    if(c<z)tmp-=z-c;
    if(tmp>=0)puts("Yes");
    else puts("No");

    return 0;
}