HDU 4000 - Fruit Ninja (树状数组 + 推理)
题意
找出按顺序且x < z < y的个数。
思路
一直想以其中的一个为起点枚举，就算知道了是树状数组也不知道树状数组怎么用TAT。还是看了题解。
原来是用树状数组统计出当前数后面有几个是比他大的。那么这时候可以组成的排列是$x * (x - 1) / 2$。但是这里面包括不符合要求的x < y < z，所以要把这部分减去。涨姿势了。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1e5 + 10;
const int MOD = 100000007;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int C[MAXN], n;
int Sum(int k)
{
int ret = 0;
while (k > 0)
{
ret += C[k];
k -= Lowbit(k);
}
return ret;
}
void Update(int k)
{
while (k <= n)
{
C[k]++;
k += Lowbit(k);
}
}
int main()
{
//ROP;
int T, i, j, cases = 0;
scanf("%d", &T);
while (T--)
{
MS(C, 0);
scanf("%d", &n);
LL ans = 0;
for (i = 1; i <= n; i++)
{
int tmp;
scanf("%d", &tmp);
Update(tmp);
LL preSmall = Sum(tmp - 1);
LL preBig = i - 1 - preSmall;
LL aftBig = n - tmp - preBig;
ans -= preSmall * aftBig;
if (aftBig >= 2) ans += aftBig * (aftBig - 1) / 2;
ans %= MOD;
}
printf("Case #%d: %lld\n", ++cases, (ans + MOD) % MOD);
}
return 0;
}