B. Different Rules
Nikolay has only recently started in competitive programming, but already qualified to the finals of one prestigious olympiad. There going to be n participants, one of whom is Nikolay. Like any good olympiad, it consists of two rounds. Tired of the traditional rules, in which the participant who solved the largest number of problems wins, the organizers came up with different rules.
Suppose in the first round participant A took x-th place and in the second round — y-th place. Then the total score of the participant A is sum x y. The overall place of the participant A is the number of participants (including A) having their total score less than or equal to the total score of A. Note, that some participants may end up having a common overall place. It is also important to note, that in both the first and the second round there were no two participants tying at a common place. In other words, for every i from 1 to n exactly one participant took i-th place in first round and exactly one participant took i-th place in second round.
Right after the end of the Olympiad, Nikolay was informed that he got x-th place in first round and y-th place in the second round. Nikolay doesn’t know the results of other participants, yet he wonders what is the minimum and maximum place he can take, if we consider the most favorable and unfavorable outcome for him. Please help Nikolay to find the answer to this question.
Input The first line contains an integer t (1≤t≤100) — the number of test cases to solve.
Each of the following t lines contains integers n, x, y (1≤n≤109, 1≤x,y≤n) — the number of participants in the olympiad, the place that Nikolay took in the first round and the place that Nikolay took in the second round.
Output Print two integers — the minimum and maximum possible overall place Nikolay could take.
Examples inputCopy 1 5 1 3 outputCopy 1 3 inputCopy 1 6 3 4 outputCopy 2 6
这个题比较简单，画一个5的所有情况，和6的所有情况，就懂了怎么做了。凑数都能凑出来。
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int t;
cin >> t;
for (int i = 0; i < t; i )
{
int a, b, c;
cin >> a >> b >> c;
if (a == 1)
cout << 1 <<" "<<1<< endl;
else{
cout << max(1, min(a, b c - a 1)) << " ";
if (b c - 1 >a)
cout << a << endl;
else
cout << min(a, a - (abs(a - b 1 - c))) << endl;
}
}
}
真的是B题题解。
末尾是个！截图没弄好，就是判断哪一些在范围内！
写在最后: 我叫风骨散人，名字的意思是我多想可以不低头的自由生活，可现实却不是这样。家境贫寒，总得向这个世界低头，所以我一直在奋斗，想改变我的命运给亲人好的生活，希望同样被生活绑架的你可以通过自己的努力改变现状，深知成年人的世界里没有容易二字。目前是一名在校大学生，预计考研，热爱编程，热爱技术，喜欢分享，知识无界，希望我的分享可以帮到你！ 如果有什么想看的，可以私信我，如果在能力范围内，我会发布相应的博文！ 感谢大家的阅读！?你的点赞、收藏、关注是对我最大的鼓励！
