zoukankan      html  css  js  c++  java
  • Codeforces Round #603 (Div. 2) A. Sweet Problem 水题

    A. Sweet Problem

    the first pile contains only red candies and there are r candies in it,
    the second pile contains only green candies and there are g candies in it,
    the third pile contains only blue candies and there are b candies in it.
    Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.

    Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.

    Input

    The first line contains integer t (1≤t≤1000) — the number of test cases in the input. Then t test cases follow.

    Each test case is given as a separate line of the input. It contains three integers r, g and b (1≤r,g,b≤108) — the number of red, green and blue candies, respectively.

    Output

    Print t integers: the i-th printed integer is the answer on the i-th test case in the input.

    Example

    input
    6
    1 1 1
    1 2 1
    4 1 1
    7 4 10
    8 1 4
    8 2 8
    output
    1
    2
    2
    10
    5
    9

    Note

    In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.

    In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.

    In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.

    题意

    给你三个数,每次你可以选择其中两个数-1,然后问你最多减多少次,使得所有数都大于等于0

    题解

    视频题解 https://www.bilibili.com/video/av77514280/

    其实答案就是min(a+b,(a+b+c)/2),考虑a+b和c的大小关系即可

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    long long a[3];
    void solve(){
    	for(int i=0;i<3;i++)
    		cin>>a[i];
    	sort(a,a+3);
    	cout<<min(a[0]+a[1],(a[0]+a[1]+a[2])/2)<<endl;
    }
    int main(){
    	int t;
    	cin>>t;
    	while(t--)solve();
    }
  • 相关阅读:
    [docker]Kubernetes的yaml文件
    [redis]redis-cluster的使用
    [redis]redis-cluster搭建
    [docker]本地仓库的创建的使用
    1W字看懂互联网知识经济
    PHP基础陷阱题(变量赋值)
    PHP不用第三变量交换2个变量的值的解决方法
    PHP中的排序函数sort、asort、rsort、krsort、ksort区别分析
    PHP实现四种基本排序算法
    WEB安全之Token浅谈
  • 原文地址:https://www.cnblogs.com/qscqesze/p/11961789.html
Copyright © 2011-2022 走看看