Codeforces Round #335 (Div. 2) A. Magic Spheres 水题-优快云博客

本文介绍了一个基于贪心算法解决的魔术球问题。该问题描述了一位魔法师如何通过转换相同颜色的球来获得所需数量的不同颜色球的过程。文章提供了完整的代码实现，并解释了背后的逻辑。

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A. Magic Spheres

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://www.codeforces.com/contest/606/problem/A

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 andz 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

Sample Output

Yes

HINT

题意

有三个物体

你现在分别有a,b,c个

你希望每种物品至少x,y,z个

然后两种相同的物品可以换成一个其他的物品

问你是否能够满足

题解：

贪心，首先，你拥有的超过需要的你才会变

然后你把这个数量记录下来，和你还差多少比一下就好了

代码：

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

int a[3];
int b[3];
int main()
{
    for(int i=0;i<3;i++)
        cin>>a[i];
    for(int i=0;i<3;i++)
        cin>>b[i];
    for(int i=0;i<3;i++)
        a[i]=a[i]-b[i];
    int flag1 = 0,flag2 = 0;
    for(int i=0;i<3;i++)
    {
        if(a[i]>0)
            flag1+=a[i]/2;
        else
            flag2+=a[i];
    }
    flag2 = -flag2;
    if(flag1>=flag2)
        printf("Yes\n");
    else
        printf("No\n");
}