Codeforces Round #603 (Div. 2)
1 second
256 megabytes
standard input
standard output
You have three piles of candies: red, green and blue candies:
- the first pile contains only red candies and there are rr candies in it,
- the second pile contains only green candies and there are gg candies in it,
- the third pile contains only blue candies and there are bb 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.
The first line contains integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Then tt test cases follow.
Each test case is given as a separate line of the input. It contains three integers rr, gg and bb (1≤r,g,b≤1081≤r,g,b≤108) — the number of red, green and blue candies, respectively.
Print tt integers: the ii-th printed integer is the answer on the ii-th test case in the input.
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
1
2
2
10
5
9
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.
题意很好理解，一行分别给出三种颜色的糖果数量。每天只能吃两个糖果，而且这两个糖果颜色不能一样。问最多能吃多少天（糖果不一定完全吃完，但是每一天都要严格按照规则）
对a,b,c先从小到大排序。
如果a+b<c，很明显，输出a+b就好了
对于a+b>=c。(a+b+c)/2这个点一定落在b上，那么我们让这个点的左右部分配对就可以了，一共就是（a+b+c)/2了，多出来也没关系。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int a[]; int sum = ;
for(int i = ; i< ;i++)
{
cin>>a[i];
sum+=a[i];
}
sort(a,a+);
if(a[]+a[]<a[])
cout<<a[]+a[]<<endl;
else
cout<<sum/<<endl;
}
}
On the well-known testing system MathForces, a draw of nn rating units is arranged. The rating will be distributed according to the following algorithm: if kk participants take part in this event, then the nn rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.
For example, if n=5n=5 and k=3k=3, then each participant will recieve an 11 rating unit, and also 22 rating units will remain unused. If n=5n=5, and k=6k=6, then none of the participants will increase their rating.
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 n=5n=5, then the answer is equal to the sequence 0,1,2,50,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋⌊n/k⌋ for some positive integer kk (where ⌊x⌋⌊x⌋ is the value of xx rounded down): 0=⌊5/7⌋0=⌊5/7⌋, 1=⌊5/5⌋1=⌊5/5⌋, 2=⌊5/2⌋2=⌊5/2⌋, 5=⌊5/1⌋5=⌊5/1⌋.
Write a program that, for a given nn, finds a sequence of all possible rating increments.
The first line contains integer number tt (1≤t≤101≤t≤10) — the number of test cases in the input. Then tt test cases follow.
Each line contains an integer nn (1≤n≤1091≤n≤109) — the total number of the rating units being drawn.
Output the answers for each of tt test cases. Each answer should be contained in two lines.
In the first line print a single integer mm — the number of different rating increment values that Vasya can get.
In the following line print mm integers in ascending order — the values of possible rating increments.
4
5
11
1
3
4
0 1 2 5
6
0 1 2 3 5 11
2
0 1
3
0 1 3
题意呢，就是给n，把其分成k份。每份就是n/k。比如样例一：
n=5: 1 2 3 4 5 6
5 2 1 1 1 0
所以输出: 0 1 2 5
我们看看n的范围,1e9，暴力的话会超时的（菜鸡的我第一发T了）我们观察一下，比如说对于样例一，3，4，5出的结果都是1，但是我们只能输出一个1，有没有办法让遍历
从n=3直接到6，跳过重复项呢？
对于样例一，5/3==1,而5/1==5,即i=5时，n恰好整除，那么我们让i==3直接跳到i==6就好了
3是第一个出现n/i==1的地方，直接让i==3跳到i==6，跳到整除的下一个i,可以跳过重复答案，因为如果n%i==0，那么n/(i+1)必定结果和n/(i)是不一样的
具体的就看代码吧，正好复习一下vector迭代器的敲法.....
#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();
for(int i=,j;i<=n;i=j+)
{
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;
}
}