uvalive4329

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2330

题目大意:一条大街住着n个乒乓爱好者,经常组织切磋技术,每个人有一个不同的技能值ai.每场比赛要3个人,两选手,一裁判.裁判要住在两选手之间,技能也是位于其之间.问能组织多少种比赛.

题目思路:假设第i个人当裁判,假设a1到ai1中有m1[i]个比ai小,那么就有(i-1)-m1[i]个比ai大.同理m2[i]表示ai+1到an中比ai小的数.那么i当裁判能组织m1[i]*(n-i-m2[i]) + (i-m1[i]-1)*m2[i]种比赛.

代码如下:

#include 
#include 

const int maxn = 100005;

int c[maxn];
int a[20005];
int m1[20005], m2[20005];

int lowbit(int x){
    return x&(-x);
}

int sum(int x){
    int ret = 0;
    while(x > 0){
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}

void add(int x, int d){
    while(x < maxn){
        c[x] += d;
        x += lowbit(x);
    }
}

int main(){
    int t, n;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            m1[i] = sum(a[i]);
            add(a[i], 1);
        }
        memset(c, 0, sizeof(c));
        for(int i = n; i >= 1; i--){
            m2[i] = sum(a[i]);
            add(a[i], 1);
        }
        long long ans = 0;
        for(int i = 1; i <= n; i++){
            ans += (long long)m1[i]*(n-i-m2[i]) + (long long)(i-m1[i]-1)*m2[i];
        }
        printf("%lld\n", ans);
    }
    return 0;
}