Codeforces Round # 603 (Div. 2) A. Sweet Problem (water ....... not do it) + C title

You have three piles of candies: red, green and blue candies:

the first pile contains only red candies and there are
the second pile contains only green candies and there are
the third pile contains only blue candies and there are

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

Each test case is given as a separate line of the input. It contains three integers

Output

Example

input

Copy

output

Copy

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.

　　A good understanding of the meaning of problems, and his party were given the number of candy in three colors. You can only eat two candy, candy colors and the two can not be the same. Q. How many days up to eat (candy may not be entirely finished, but every day should be strictly in accordance with the rules)

　　For a, b, c first from small to large.

　　If a + b <c, it is clear that, like the output of a + b

　　For a + b> = c. (A + b + c) / 2 this point must fall on b, then we make the point about pairing part of it, is a total of (a + b + c) / 2, and more out of it does not matter.

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std; 
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int a[3];
        
        int sum = 0;
        for(int i = 0 ; i<3 ;i++)
        {
            cin>>a[i];
            sum+=a[i];
        }
        sort(a,a+3);
        if(a[0]+a[1]<a[2])
            cout<<a[0]+a[1]<<endl;
        else
            cout<<sum/2<<endl;
    }
}

　　　　　　C_Everyone is a Winner!

　　　　　　On the well-known testing system MathForces, a draw of

　　　　　　For example, if

　　　　　　Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.

　　　　　　For example, if

　　　　　　Write a program that, for a given

Input

The first line contains integer number

Each line contains an integer

Output

Output the answers for each of

In the first line print a single integer

In the following line print

Example

input

Copy

output

Copy

. 4 
0 2. 1. 5 
. 6 
0 2. 3. 5. 11. 1 
2 
0. 1 
. 3 
0. 1. 3 
　　　　　meaning of problems, that is, to n, which is divided into parts to k. Each is n / k. For example, a sample: 
　　　　n = 5: 1 2 3 4 5 6 
　　　　　　　　　　　　 5 2 1 1 1 0 
　　　　0 1 2 5: the output 
　　　　I made our first look at the range of n, 1e9, violent words will timeout (chicken dish T a) we look, for example, for a sample, the results are 1, 4, 5, but we can only output a 1, there is no way to traverse 
　from n = 3 to 6 directly, skipping repeat item it? 
　　　　For a sample, 5/3 == 1 and 5/1 == 5, i.e., i = 5, n is exactly divisible, then we let I 3 == skip to i == 6 like 
　　　　3 are a place n / i == 1 occurs, skip directly to i == 3 i == 6, next a jump i divisible, can skip repeat answer because if n% i == 0, then n / (i + 1) and the results must be n / (i) is not the same 
　　　　specific codes to see it, just review the vector iterator knock law .....

#include<iostream>
#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int t; 
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        vector<int>vv;
        vv.push_back(0);
        for(int i=1,j;i<=n;i=j+1)
        {
            vv.push_back(n/i);
            j=n/(n/i);
        }
        sort(vv.begin(),vv.end());
        cout<<vv.size()<<endl;
        vector<int>::iterator i;    //vector迭代器 
        for(i=vv.begin();i<vv.end();i++)
        {
            cout<<*i<<" ";
        }
        cout<<endl;
    }
}

Codeforces Round # 603 (Div. 2) A. Sweet Problem (water ....... not do it) + C title

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