P1966 火柴排队

emmmmm不难吧。。

主要就是在最开始要把题目抽象化，转为求逆序对个数，那之后就很简单了，离散化+归并排序求逆序对，取模输出就结束了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<algorithm>
#define ll long long
using namespace std;
const int oo=0x3f3f3f3f;
const int N=1000005;
const int mod=99999997;

int n,ans;
int f[N],f1[N];
struct node{
    int x,y;
}a[N];
struct node1{
    int x,y;
}b[N];

bool cmp1(node a,node b){return a.x<b.x;}
bool cmp2(node1 a,node1 b){return a.x<b.x;}

ll get(){
    char zy=getchar();
    ll z=1,y=0;
    while(zy>'9'||zy<'0'){
        if(zy=='-') z=-1;
        zy=getchar();
    }
    while(zy>='0'&&zy<='9'){
        y=(y<<1)+(y<<3)+zy-'0';
        zy=getchar();
    }
    return z*y;
}

void merge_sort(int l,int r){
    if(l>=r) return;
    int mid=(l+r)>>1;
    merge_sort(l,mid);
    merge_sort(mid+1,r);
    int p=l,q=mid+1,g=l;
    while(p<=mid||q<=r){
        if(q>r||(f[p]<=f[q]&&p<=mid)) f1[g++]=f[p++];
        else{
            ans+=mid-p+1;
            ans%=mod;
            f1[g++]=f[q++];
        }
    }
    for(int i=l;i<=r;i++){
        f[i]=f1[i];
    }
}

int main(){
    //freopen("P1966火柴排队.in","r",stdin);
    //freopen("P1966火柴排队.out","w",stdout);
    n=get();
    for(int i=1;i<=n;i++){
        a[i].x=get();a[i].y=i;
    }
    sort(a+1,a+1+n,cmp1);
    for(int i=1;i<=n;i++){
        b[i].x=get();b[i].y=i;
    }
    sort(b+1,b+1+n,cmp2);
    for(int i=1;i<=n;i++){
        f[a[i].y]=b[i].y;
    }
    for(int i=1;i<=n;i++){
    //    printf("%d ",f[i]);
    }
    merge_sort(1,n);
    printf("%d
",ans%mod);
    return 0;
}