HDU 5592 ZYB's Permutation(树状数组)

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题意：给出i位置的逆序对数前缀和，求原序列。

思路：从n开始往前推，树状数组维护已经出现的比i大的有几个，然后二分找。O(nlognlogn)。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 5e4 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int sum[N], a[N], b[N], bit[N];int cal(int i) {    int s = 0;    while(i > 0) {        s += bit[i];        i -= i & -i;    }    return s;}void add(int i, int x) {    while(i <= n) {        bit[i] += x;        i += i & -i;    }}int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);    //freopen("/Users/apple/out.txt", "w", stdout);#endif        si(T);    while(T --) {        si(n);        mem(bit, 0);        sum[0] = a[0] = 0;        for(int i = 1; i <= n; i ++) {            si(sum[i]);            a[i] = sum[i] - sum[i-1];        }        for(int i = n; i >= 1; i --) {            int l = 0, r = n + 1, mid;            while(l < r - 1) {                mid = (l + r) >> 1;                int cnt = cal(n + 1 - mid);                int tmp = n - cnt - a[i];                if(tmp < mid) r = mid;                else l = mid;            }            b[i] = l;            add(n + 1 - l, 1);        }        for(int i = 1; i < n; i ++) printf("%d ", b[i]);        printf("%d\n", b[n]);            }        return 0;}

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