描述
题解
这个题看着很复杂,代码却是十分简单,直接看代码吧,不是特别难。几个判断就能搞定了……
代码
#include <iostream>
using namespace std;
typedef long long ll;
ll n, m;
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%lld%lld", &n, &m);
if (m >= n - 1)
{
ll sum = n * (n - 1) / 2;
if (m > sum)
{
printf("%lld\n", sum * 2);
}
else
{
ll ans = (sum - m) * 2ll + m;
printf("%lld\n", ans * 2);
}
}
else
{
ll sum = n * (n - 1) / 2;
ll ans = m + m * (m - 1) + n * 1ll * (sum - m - m * (m - 1) / 2);
printf("%lld\n", ans * 2);
}
}
return 0;
}