【CodeForces 606A】A -特别水的题1-Magic Spheres
【CodeForces 606A】A -特别水的题1-Magic Spheres
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#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 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 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
这题就是说两个相同的球可以变成一个任意一种球，现在问你能否得到各种球至少给定个？用当前的个数减去目标个数，如果是正数，多出来的可以拿去变成其它的，于是可以得到[差除以二]个其它球（整除），我们就ans+=差/2。如果是负数，就是需要变成[-差]那么多个，就ans+=差。最后ans是负数，则不能，否则能。
#include<stdio.h>
int main()
{
long long a,b,c,x,y,z,ans=0;
scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&x,&y,&z);
a=a-x;
b=b-y;
c=c-z;
if(a>0)ans+=a/2;
else ans+=a;
if(b>0)ans+=b/2;
else ans+=b;
if(c>0)ans+=c/2;
else ans+=c;
if(ans>=0)printf("Yes");
else printf("No");
return 0;
}