2020 cattle off winter training camp algorithm base 6 D rearrangement

https://ac.nowcoder.com/acm/contest/3007/D

 

Although the obtained solution to a problem with the same conclusion, but still more than problem solution thinking trouble. .

10 questions over Question 8, 6 questions do trouble

Vomiting blood. . .

 

The A and B are sorted in ascending

A number of position then each is able to fill a range, the range to the last position is the right end, left end point does not fall

My idea is from a large number of small to fill, so that each can fill the position will not increase

A one bit for each selected to cover all of B in their own section in

F_n [i] represents the n bits, and now to the i-th

Because we can put the number of the i-th bit will be able to put the first bit i + 1, the i + 1 corresponding to the first consider bit away when the i-th bit into a

f_n [i] = f_n-1 [i + 1] * L [i], L [i] denotes the i-th bit interval coverage is much

f_n-1[i+1]=f_n-2[i+2]*(L[i+1]-1)

f_n-2[i+2]=f_n-3[i+3]*(L[i+2]-2)

……

f_n[1]=L[1]*(L[2]-1)*(L[3]-2)*(L[4]-3)*(L[5]-4)*(L[n]-(n-1))

 

Problem solutions from small to large to fill

Because the smaller that position need to have a certain number of fill

The i-th position can be placed into the first must be able to position i + 1

Therefore, the program number = L [. 1] * (L [2] -1) * (L [. 3] -2) * (L [. 4] -3) * (L [. 5] -4) * (L [n-] - (n-1))

 

 

#include<cstdio>
#include<algorithm>
 
using namespace std;
 
#define N 100001
 
const int mod=1e9+7;
 
int a[N],b[N];
int L[N];
 
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    for(int i=1;i<=n;++i) scanf("%d",&b[i]);
    sort(b+1,b+n+1);
    sort(a+1,a+n+1);
    int m=n;
    for(int i=n;i;--i)
    {
        while(m && a[i]<=b[m]) m--;
        L[m+1]++;
        if(i<=m)
        {
            printf("0");
            return 0;
        }
    }
    for(int i=2;i<=n;++i) L[i]+=L[i-1];
    int ans=1;
    for(int i=1;i<n;++i) ans=1ll*ans*(L[i]-i+1)%mod;
    printf("%d",ans);
    return 0;
}