feat: 添加P5200题解实现，使用树状数组计算逆序对

实现了一个使用树状数组计算逆序对数量的算法，用于解决P5200题目。通过维护树状数组来高效计算区间和，从而确定每个元素的逆序对数量。

src/8/31/P5200.cpp

@@ -0,0 +1,65 @@
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <vector>
+using ll = int64_t;
+
+ll n;
+std::vector<ll> v,t,ans;
+
+#define pv(v)do{std::cout<<#v<<" :"<<(v)<<"\n";}while(0)
+
+static inline constexpr ll lowbit(ll n){
+    return n&(-n);
+}
+
+static inline void set(ll idx,ll addval){
+    while(idx<=n){
+        t[idx]+=addval;
+        idx+=lowbit(idx);
+    }
+}
+
+static inline ll getsfx(ll idx){
+    ll res=0;
+    while(idx){
+        res+=t[idx];
+        idx-=lowbit(idx);
+    }
+    return res;
+}
+
+static inline ll getrange(ll l,ll r){
+    return getsfx(r)-getsfx(l-1);
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n;
+    v.resize(n+1);
+    t.resize(n+1);
+    ans.reserve(n+1);
+    for(ll i=1;i<=n;++i){
+        std::cin>>v[i];
+    }
+    ll idx=n;
+    for(;idx>=2 && v[idx-1]<=v[idx];--idx);
+    // pv(idx);
+    for(ll i=idx;i<=n;i++){
+        set(v[i], 1);
+    }
+    ll sum=0;
+    for(ll i=1;i<idx;i++){
+        ll add = idx-i-1 + getrange(1, v[i]);
+        sum+=add;
+        ans.push_back(add);
+        set(v[i], 1);
+    }
+    std::cout<<idx-1<<"\n";
+    for(ll i:ans){
+        std::cout<<i<<" ";
+    }
+    std::cout<<"\n";
+}