# Ultraviolet A 1428-Ping pong (tree array)

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Ultraviolet A 1428-Ping pong (tree array)

Link to the question: Ultraviolet A 1428-Ping pong

There are n table tennis fans living in a street and they often organize competitions. Each person has a different ability value. Each game requires three people. The referee must live between two players and the ability value must also be between players, ask how many matches can be held at most.

Solution: Pre-processing bi and ci are represented in 1 ~ In I, the energy value is smaller than that in I ~ The energy value in n is smaller than that in I. You can use a tree array to maintain the processing process.

``````#include

#include

#include using namespace std;#define lowbit(x) ((x)&(-x))const int maxn = 1e5;typedef long long ll;int n, s[maxn+5];int N, a[maxn+5];ll b[maxn+5];void add (int x, int v) {    while (x <= n) {        s[x] += v;        x += lowbit(x);    }}int sum (int x) {    int ret = 0;    while (x > 0) {        ret += s[x];        x -= lowbit(x);    }    return ret;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &N);        n = 0;        memset(s, 0, sizeof(s));        for (int i = 1; i <= N; i++) {            scanf("%d", &a[i]);            n = max(a[i], n);        }        for (int i = 1; i <= N; i++) {            b[i] = sum(a[i]-1);            //printf("%lld ", b[i]);            add(a[i], 1);        }        //printf("\n");        ll ans = 0;        memset(s, 0, sizeof(s));        for (int i = N; i > 0; i--) {            ll d = sum(a[i]-1);            add(a[i], 1);            ans += (b[i] * (N - i - d)) + d * (i - 1 - b[i]);        }        printf("%lld\n", ans);    }    return 0;}

``````

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