for a give n x we are calculating all permutations v = all value less then x we can always put all v values in first place x could always defeat these values and will become x+1 now in second position e can select v -1 as out of v we have already put 1 value at first place but, we can also put those values which are equal to x+1 thats why for x+1 we added v += the num of enemy that has exactly x+1 candies. so we should multiply our ans with this num thatis ans = v*(v-1 + k) where k is num of enemies that have exactly same candies as x+1 and v is numbver of candies less then x, and so on.

https://codeforces.me/blog/himanshu_attri Hello all my brothers it will be very kind of you if you can help me solving with this problem im stuck with this problem comes in teksystem hiring round and i was unable to solve it please dont unvote if you dont want to help let somebody else do it thank you very much

https://codeforces.me/blog/himanshu_attri please someone help me with this problem it will be very helpful thank you in advance this is first time i am asking for help

https://codeforces.me/blog/himanshu_attri please someone help me with this problem i stuck with

in c why we initialise res=m? why can't we add res+=1 in the else statement when pos[b[i]] < last ??

in editorial of C2 why if (vals.back() == 1) ans = pw / 3; we are doing this