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题目链接注意：每次求逆序或正序和都需要初始化#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#i

题目链接思路类似于校门外的树那道题#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;typedef long long ll;const int N=100010;int c1[N],c2[N];int...

题目链接树状数组模板题#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;typedef long long ll;const int N=500010;int c[N];int n,m;int low...

题目链接将区间修改转化为两个单点修改，则查询操作的结果为r左边的（个数减去l左边的）的个数，进而想到树状维护前缀和，具体思路是c1维护（的个数，c2维护）的个数，#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std...

题目链接y从小到大排序，y相同x从小到大排序，转化为当前星星i之前，x小于等于的i的星星个数，即为这个星星的级数（正序对问题）#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;typedef long lon...

题目链接树状数组求逆序对#include<iostream>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int N=200010;int n;int a[N],r[N],l[N],c[N];int ask(int x){...

题目链接#include <cstring>#include <cstdlib>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <map>#include <stack&a

题目链接直接二维树状数组会超内存，利用贪心排序降维（题目数据已经有序）#include <cstring>#include <cstdlib>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <...

题目链接1.利用等差数列转化为求逆序对和正序对2.树状数组优化因为ans的输出问题WA了好多次。。#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const ...