codechef LEPERMUT - "little elephant and permutations" solution and guidance

codechef LEPERMUT - "little elephant and permutations": http://www.codechef.com/problems/LEPERMUT

codechef LEPERMUT - "little elephant and permutations" solution: http://ideone.com/zSwh4P

#include <iostream>

#include <cstdio>

using namespace std;

int main() {

int t, n, a[105], n_inv, l_inv;

scanf("%d", &t);

while(t--) {

scanf("%d", &n);

n_inv=l_inv=0;

for(int i=0; i<n; i++) scanf("%d", &a[i]);

for(int i=0; i<n-1; i++) for(int j=i+1; j<n; j++) if(a[i]>a[j]) n_inv++;

for(int i=1; i<n; i++) if(a[i-1]>a[i]) l_inv++;

if(l_inv==n_inv) printf("YES\n");

else printf("NO\n");

}

return 0;

}

codechef LEPERMUT - "little elephant and permutations" guidance:

number inverse is all inverse in the loop; however, local inverse is, as name also indicates, consecutive number inverse.


thank God, i am happy with my progression :)