UVALive 4329 Ping pong

树状数组。考虑ai（从0开始，则i左边共i个，右边n-i-1个），左边有x个比他大的，i-x个比他小的，右边有y个比他大的，n-i-1-y个比他大的。交叉乘一下就得到了以ai为裁判的比赛总数。把所有人都枚举一遍，加在一起就是答案，会超int。

如何才能知道ai左边有多少比他小的呢？假如aj<ai且j<i，用另一个数组b来标记这个数有没有出现过，那么b[aj] = 1;这样，比ai小的数的个数，就是b的前缀和。

 
#include<algorithm>  
#include<iostream>  
#include<cstring>  
#include<cstdio>  
#include<vector>  
#include<string>  
#include<queue>  
#include<cmath>  
///LOOP  
#define REP(i, n) for(int i = 0; i < n; i++)  
#define FF(i, a, b) for(int i = a; i < b; i++)  
#define FFF(i, a, b) for(int i = a; i <= b; i++)  
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)  
#define FDD(i, a, b) for(int i = a; i >= b; i--)  
///INPUT  
#define RI(n) scanf("%d", &n)  
#define RII(n, m) scanf("%d%d", &n, &m)  
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)  
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)  
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)  
#define RFI(n) scanf("%lf", &n)  
#define RFII(n, m) scanf("%lf%lf", &n, &m)  
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)  
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)  
#define RS(s) scanf("%s", s)  
///OUTPUT  
#define PN printf("\n")  
#define PI(n) printf("%d\n", n)  
#define PIS(n) printf("%d ", n)  
#define PS(s) printf("%s\n", s)  
#define PSS(s) printf("%s ", n)  
///OTHER  
#define PB(x) push_back(x)  
#define CLR(a, b) memset(a, b, sizeof(a))  
#define CPY(a, b) memcpy(a, b, sizeof(b))  
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}  
  
using namespace std;  
typedef long long LL;  
typedef pair<int, int> P;  
const int MOD = 100000000;  
const int INFI = 1e9 * 2;  
const LL LINFI = 1e17;  
const double eps = 1e-6;  
const double pi = acos(-1.0);  
const int N = 111111;  
const int M = 22;  
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};  
  
int a[N], b[N], c[N], n;  
  
void update(int x)  
{  
    while(x <= N)  
    {  
        b[x]++;  
        x += x & (-x);  
    }  
}  
  
int sum(int x)  
{  
    LL s = 0;  
    while(x)  
    {  
        s += b[x];  
        x -= x & (-x);  
    }  
    return s;  
}  
  
int main()  
{  
    //freopen("input.txt", "r", stdin);  
    //freopen("output.txt", "w", stdout);  
  
    int t, k;  
    LL ans;  
    RI(t);  
    while(t--)  
    {  
        RI(n);  
        CLR(b, 0);  
        REP(i, n)  
        {  
            RI(a[i]);  
            c[i] = sum(a[i]);  
            update(a[i]);  
        }  
        CLR(b, 0);  
        ans = 0;  
        FD(i, n, 0)  
        {  
            k = sum(a[i]);  
            ans += LL(c[i]) * LL(n - i - 1 - k) + LL(i - c[i]) * LL(k);  
            update(a[i]);  
        }  
        printf("%lld\n", ans);  
    }  
    return 0;  
}

补充：软件开发 , C++ ,