P1966 火柴排队

P1966 火柴排队
进行映射，然后求逆序对即可。

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define For(i,a,b) for(register int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()
#define mod 99999997
//by war
//2017.11.8
using namespace std;
int n;
int a[100010];
int b[100010];
int c[100010];
int t[100010];
int ans;
void in(int &x)
{
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9')
    {
    if(c=='-')
    y=-1;
    c=g();
    }
    while(c<='9'&&c>='0')x=(x<<1)+(x<<3)+c-'0',c=g();
    x*=y;
}
void o(int x)
{
    if(x<0)
    {
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}

void modify(int k)
{
    for(;k<=n;k+=(-k)&k)
    t[k]++;
}

int get(int k)
{
    int cnt=0;
    for(;k>0;k-=(-k)&k)
    cnt=(cnt+t[k])%mod;
    return cnt;
}

int main()
{
    in(n);
    For(i,1,n)
    in(a[i]),c[a[i]]=i;
    For(i,1,n)
    in(b[i]),b[i]=c[b[i]];
    For(i,1,n)
    {
        if(b[i]!=0)
        {
        modify(b[i]);
        ans=(ans+i-get(b[i]))%mod;    
        }
        else
        {
        modify(n);
        ans=(ans+i-get(n))%mod;    
        }
    }
    o(ans%mod);
     return 0;
}