2025. Line Fighting
Time limit: 1.0 second
Memory limit: 64 MB
Memory limit: 64 MB
Boxing, karate, sambo… The audience is sick of classic combat sports. That is why a popular sports channel launches a new competition format based on the traditional Russian entertainment called line fighting.There can be from
2 to k teams taking part in a competition, and there are n fighters altogether in all the teams. Before the competition starts, the fighters are divided into teams: each fighter becomes a member of exactly one team.Two fighters fight each
other if they are members of different teams. The organizers believe that the more the number of fights between fighters, the higher the popularity of a competition will be. Help the organizers to distribute fighters between teams so as to maximize the number
of fights and output this number.
Input
The first line contains the number of tests T (1 ≤
T ≤ 10). In each of the following T lines you are given a test:integers
n and k separated with a space (2 ≤ k ≤ n ≤ 104).
Output
For each test output the answer (one integer) in a separate line.
Sample
| input | output |
|---|---|
3 6 3 5 5 4 2 |
12 10 4 |
Problem Source: Ural Regional School Programming Contest 2014
解析：组合数学。因为组内不能打比赛，这就相当于在全部人都能比赛的基础上去掉了各个组间能打的比赛次数。
首先，比赛次数最多的情况肯定是尽可能地将人数均分，这种比赛数是最多的。
AC代码：
#include <bits/stdc++.h>
using namespace std;
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif // sxk
int T, n, k, ans;
scanf("%d", &T);
while(T --){
scanf("%d%d", &n, &k);
ans = n * (n - 1) / 2; //全部人两两之间打比赛的次数
if(n != k){
int foo = n / k;
int cnt = n % k; //均分后剩余cnt个人，再均分。则会出现cnt个人数多1的组
ans -= cnt * ((foo + 1) * foo / 2); //去掉人数较多的cnt组的总次数
ans -= (k - cnt) * (foo * (foo - 1) / 2); //去掉人数较少的总次数
}
printf("%d
", ans);
}
return 0;
}