Codeforces Round #335 (Div. 2) A. Magic Spheres 水题

最新推荐文章于 2017-01-15 17:00:27 发布

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最新推荐文章于 2017-01-15 17:00:27 发布

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分类专栏： Codeforces

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本文介绍了一个关于魔法球颜色转换的问题，魔术师需要通过转换相同颜色的球来获得不同颜色的球，以达到特定的目标数量。文章提供了问题背景、输入输出格式及样例，并附带了解决该问题的代码实现。

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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 test(s)

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

题意：有三种颜色数量分别是a、b、c，然后每两种相同的颜色可以转换成其他的一种颜色，如果颜色的数量分别能大于等于x、y、z，输出yes，否则输出no。

思路：存在某种颜色不足够，用sum的值去保存，等待转换，否则就把多余的记为（p）转换，即sum += p/2。

代码：

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
#define maxn 100000+50
int main()
{
#ifdef CDZSC_June
    freopen("t.txt","r",stdin);
#endif
    int a[4],b[4],sum = 0;
    scanf("%d%d%d",&a[0],&a[1],&a[2]);
    scanf("%d%d%d",&b[0],&b[1],&b[2]);
    for(int i = 0; i<3; i++)
    {
        if((a[i] - b[i]) < 0)
        {
            sum += a[i] - b[i];
        }
        else
        {
            sum += (a[i] - b[i])/2;
        }
    }
    puts(sum >= 0?"Yes":"No");
}