Uva1315-crazy Tea Party (derivation)

Question Link

N people sit in a ring, and the adjacent two can exchange positions. The minimum number of exchanges leads to the opposite of the sequence.

Idea: similar to Bubble sorting, the circular sequence can be split into two sequences for bubble respectively. When N is an odd number, it is divided into n/2 and N/2 + 1, so ans = (n/2) * (N/2-1) /2 + (n/2) * (N/2 + 1)/2. When n is an even number, it is divided into two n/2, so ans = (n/2) * (N/2-1 ).

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n;int main() {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &n);            if (n == 1 || n == 2)              printf("0\n");        else {            int ans;            if (n % 2 == 0) {                ans = (n / 2) * (n / 2 - 1);                printf("%d\n", ans);            }            else {                ans = (n / 2) * (n / 2 - 1) / 2 + (n / 2) * (n / 2 + 1) / 2;                printf("%d\n", ans);             }         }     }    return 0;}

Uva1315-crazy Tea Party (derivation)