牛客多校四

c题晚了一分钟交就不再多写
i题 Inverse Pair
比赛的时候不会归并排序减小复杂度,补个归并排序的板子

void merge_sort(int start,int end){
	if(start>=end){
		return;
	}
	int start1=start,end1=(start+end)/2,start2=end1+1,end2=end;
	merge_sort(start1,end1);
	merge_sort(start2,end2);
	int k=start;
	while(start1<=end1&&start2<=end2){
		arr[k++]=a[start1]<a[start2]?a[start1++]:a[start2++];
	}
	while(start1<=end1){
		arr[k++]=a[start1++];
	}
	while(start2<=end2){
		arr[k++]=a[start2++];
	}
	for(int i=start;i<=end;i++){
		a[i]=arr[i];
	}
}

此题就是对于数列中当前的x,如果存在值x+1在它之前则将它加一,得出数列利用归并排序得出逆序对数

#include<bits/stdc++.h>
using namespace std;

#define int long long
const int N=2e5+7;
int n,ans;
int a[N],arr[N];
map<int,int> mp;
void merge_sort(int start,int end){
	if(start>=end){
		return;
	}
	int start1=start,end1=(start+end)/2,start2=end1+1,end2=end;
	merge_sort(start1,end1);
	merge_sort(start2,end2);
	int k=start;
	int p=start2;
	while(start1<=end1&&start2<=end2){
		if(a[start1]>a[start2])ans+=p-start1;
		arr[k++]=a[start1]<=a[start2]?a[start1++]:a[start2++];
	}
	while(start1<=end1){
		arr[k++]=a[start1++];
	}
	while(start2<=end2){
		arr[k++]=a[start2++];
	}
	for(int i=start;i<=end;i++){
		a[i]=arr[i];
	}
}
signed main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]==1){
        	a[i]++;
        	continue;
		};
        if(a[i]==n){
			mp[a[i]]++;
        	continue;
		}  
        if(mp[a[i]+1]>0)a[i]++;
        mp[a[i]]++;    
        
    }
    merge_sort(1,n);
	printf("%lld\n",ans);
	return 0;
}