[hdu4911]逆序对相关

思路：由于只能交换相邻的数，所以每次最多减小1个逆序对（且如果存在逆序对那么肯定可以减小1个）！于是乎。。就是统计逆序对的裸题了。树状数组或归并都行。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define Rep(a, b) for(int a = 0; a < b; a++)
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef multiset<int>::iterator msii;
typedef set<int> si;
typedef set<int>::iterator sii;
typedef vector<int> vi;

const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
const int maxn = 1e5 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-10;

int tmp[maxn], a[maxn];

LL merge(int l, int m, int r) {
    int p = m + 1, t = l;
    LL res = 0;
    for (int i = l; i <= m; i++) {
        while (p <= r && a[i] > a[p]) {
            tmp[t++] = a[p++];
        }
        res += p - m - 1;
        tmp[t++] = a[i];
    }
    for (int i = l; i < p; i++) a[i] = tmp[i];
    return res;
}

LL merge_sort(int l, int r) {
    if (l == r) return 0;
    define_m;
    return merge_sort(l, m) + merge_sort(m + 1, r) + merge(l, m, r);
}

int main() {
    //freopen("in.txt", "r", stdin);
    int n;
    LL k;
    while (cin >> n >> k) {
        for (int i = 0; i < n; i++) {
            scanf("%d", a + i);
        }
        LL ans = merge_sort(0, n - 1) - k;
        ans = max(ans, 0LL);
        cout << ans << endl;
    }
    return 0;
}

Hint:做这题的时候发现一个不容易发现的坑（与这题本身无关），多个函数相加并不一定按从左至右的顺序执行（可能是为了某种意义上地优化代码）。此代码用c++交好像w会a掉，不信可以一试~(c++貌似优化程度比较高？)