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)?
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.
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
4 4 0
2 1 2
Yes
5 6 1
2 7 2
No
3 3 3
2 2 2
Yes
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.
题意：给你三种球数量分别是a,b,c，对于两个相同的球可以转化为任意一个其他球，现在问你能否得到至少x,y,z三种球的数量
题解：
对于a-x,大于0，则价值为(a-x)/2;小于0，则价值为(a-x);b,c类似，最后判断价值是否为正就好了
//meek
///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
//**************************************** const int N=+;
const ll inf = 1ll<<;
const int mod= ; int main() {
int a,b,c,x,y,z;
scanf("%d%d%d",&a,&b,&c);
scanf("%d%d%d",&x,&y,&z);
int aa=a-x;
int bb=b-y;
int cc=c-z;
ll tmp=;
if(aa>) {
tmp+=(aa/);
}
else tmp+=(aa);
if(bb>) {
tmp+=(bb/);
}
else{
tmp+=(bb);
}
if(cc>) {
tmp+=(cc/);
}
else tmp+=(cc);
if(tmp>=) {
cout<<"Yes"<<endl;
}
else cout<<"No"<<endl; return ;
}
代码