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

    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 ab 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, xy 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
    ");
        else
            printf("No
    ");
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5036324.html
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