Nowcoder -- Niuke lefa red packet off single ACM competition C _ zone Interval

Link to the question: Inter-zone C _

Idea: Calculate contribution and find the number of intervals where each number is the current maximum and the number of intervals where each number is the minimum.

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# Include <bits/stdc ++. h> using namespace STD; # define maxn 100005 # define ll long longll ll [maxn], RR [maxn], a [maxn]; ll work (ll n) {memset (LL, 0, sizeof (LL); memset (RR, 0, sizeof (RR); stack <int> S, T; For (LL j = 1; j <= N; j ++) {While (S. size () & A [J]> = A [S. top ()]) {S. pop () ;}if (! S. size () ll [J] = 1; else ll [J] = S. top () + 1; S. push (j) ;}for (ll j = N; j> = 1; j --) {While (T. size () & A [J]> A [T. top ()]) {T. pop () ;}if (! T. size () rr [J] = N; else RR [J] = T. top ()-1; T. push (j);} ll ans = 0; For (LL j = 1; j <= N; j ++) {ans + = 1ll * A [J] * 1ll * (RR [J]-ll [J] + (RR [J]-j) * (J-ll [J]);} return ans;} int main () {ll t; CIN> T; while (t --) {ll N; cin> N; For (LL j = 1; j <= N; j ++) {scanf ("% d", & A [J]);} ll ans = work (n); // cout <ans <Endl; For (LL j = 1; j <= N; j ++) {// when the value is assigned, the result equivalent to finding the maximum value is a negative number, which is equal to minus a [J] =-A [J];} ans + = work (N ); cout <ans <Endl ;}}

 

Nowcoder -- Niuke lefa red packet off single ACM competition C _ zone Interval