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  • CodeForces--606A --Magic Spheres(模拟水题)

    Time Limit: 2000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u

    Status

    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
    

    Sample Output

    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.

    Source


    现在给你a,b,c个不同颜色的球,神魔颜色我也不知道,两个相同颜色的可以得到一个任意颜色的,问现在是否可以得到x,y,z,个三种颜色的球
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int a,b,c,x,y,z;
    long long ans,pre;
    int main()
    {
    	while(scanf("%d%d%d%d%d%d",&a,&b,&c,&x,&y,&z)!=EOF)
    	{
    		ans=pre=0;
    		int p,q,r;
    		p=q=r=0;
    		if(a-x>=0) p=0,pre+=(a-x)/2;
    		else p=a-x;
    		if(b-y>=0) q=0,pre+=(b-y)/2;
    		else q=b-y;
    		if(c-z>=0) r=0,pre+=(c-z)/2;
    		else r=c-z;
    //		printf("%d
    ",pre);
    //		printf("%d %d %d
    ",p,q,r);
    		if(pre+p+q+r>=0) printf("Yes
    ");
    		else printf("No
    ");
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273522.html
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