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wvp平台 视频设备接入注意点

一、第一题:火柴排队

解题思路:归并排序
        重点在于想清楚是对哪个数组进行归并排序求逆序对

【Python程序代码】

from math import *
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
na,nb = [],[]
for i in range(n):
    na.append([a[i],i])
    nb.append([b[i],i])
na.sort()
nb.sort()
np = [0]*(n+5)
for i in range(n):
    np[na[i][1]] = nb[i][1]
tep = [0]*(100005)
def merge_sort(q, l, r):
    if l >= r: return 0
    mid = (l + r) >> 1
    res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r)
    k, i, j = 0, l, mid + 1
    while i <= mid and j <= r:
        if q[i] <= q[j]:
            tep[k] = q[i]
            k, i = k + 1, i + 1
        else:
            res += mid - i + 1
            tep[k] = q[j]
            k, j = k + 1, j + 1

    while i <= mid:
        tep[k] = q[i]
        k, i = k + 1, i + 1
    while j <= r:
        tep[k] = q[j]
        k, j = k + 1, j + 1
    j = 0
    for i in range(l, r + 1):
        q[i] = tep[j]
        j += 1
    return res
print( (merge_sort(np,0,n-1))%99999997)

二、第二题: 归并排序

解题思路:归并排序
        归并排序模板题

【Python程序代码】

n = int(input())
a = list(map(int,input().split()))
tep = [0]*(n+5)
def merge_sort(q,l,r):
    if l>=r:return
    mid = (l+r)>>1
    merge_sort(q,l,mid);merge_sort(q,mid+1,r)
    k,i,j = 0,l,mid+1
    while i<=mid and j<=r:
        if q[i]<=q[j]:
            tep[k] = q[i]
            k,i=k+1,i+1
        else:
            tep[k] = q[j]
            k,j=k+1,j+1
    while i<=mid:
        tep[k]=q[i]
        k,i=k+1,i+1
    while j<=r:
        tep[k]=q[j]
        k,j=k+1,j+1
    j = 0
    for i in range(l,r+1):
        q[i]=tep[j]
        j += 1

merge_sort(a,0,n-1)
for i in range(n):
    print(a[i],end=" ")

三、第三题:逆序对的数量

解题思路:归并排序
        归并排序求逆序对模板题

【Python程序代码】

n = int(input())
a = list(map(int,input().split()))
tep = [0]*(n+5)
def merge_sort(q,l,r):
    if l>=r:return 0
    mid = (l+r)>>1
    res = merge_sort(q,l,mid)+merge_sort(q,mid+1,r)
    k,i,j = 0,l,mid+1
    while i<=mid and j<=r:
        if q[i]<=q[j]:
            tep[k] = q[i]
            k,i=k+1,i+1
        else:
            res += mid - i + 1
            tep[k] = q[j]
            k,j=k+1,j+1
            
    while i<=mid:
        tep[k]=q[i]
        k,i=k+1,i+1
    while j<=r:
        tep[k]=q[j]
        k,j=k+1,j+1
    j = 0
    for i in range(l,r+1):
        q[i]=tep[j]
        j += 1
    return res
print(merge_sort(a,0,n-1))

 四、第四题:小朋友排队

解题思路:归并排序
        归并排序求出每个数与其他数组成的逆序对数,然后求和公式累加

【Python程序代码】

n = int(input())
a = list(map(int, input().split()))
q,tep = [],[[0]*2 for _ in range(n+5) ]
for i in range(n):q.append( [a[i],i] )
sum = [0]*(n+5)
def merge(q,l,r):
    if l>=r:return
    mid = (l+r)>>1
    merge(q,l,mid);merge(q,mid+1,r)
    k,i,j=0,l,mid+1
    while i<=mid and j<=r:
        if q[i][0]<=q[j][0]:
            tep[k]=q[i]
            sum[q[i][1]] += j-mid-1
            k,i = k+1,i+1
        else:
            tep[k]=q[j]
            sum[q[j][1]] += mid-i+1
            k,j = k+1,j+1
    while i<=mid:
        tep[k] = q[i]
        sum[q[i][1]] += j-mid-1
        k,i = k+1,i+1
    while j<=r:
        tep[k] = q[j]
        k,j = k+1,j+1
    j = 0
    for i in range(l,r+1):
        q[i]=tep[j]
        j += 1
merge(q,0,n-1)
res = 0
for i in range(n):
    res += (sum[i])*(sum[i]+1)//2
print(res)



五、第五题:超快速排序

解题思路:归并排序
        归并排序求逆序对模板题

【Python程序代码】

import sys
tep = [0] * (500005)
def merge_sort(q,l,r):
    if l>=r:return 0
    mid = (l+r)//2
    res = merge_sort(q,l,mid) + merge_sort(q,mid+1,r)
    k,i,j = 0,l,mid+1
    while i<=mid and j<=r:
        if q[i]<=q[j]:
            tep[k]=q[i]
            k,i = k+1,i+1
        else:
            tep[k]=q[j]
            res += mid-i+1
            k,j = k+1,j+1
    while i<=mid:
        tep[k] = q[i]
        k,i = k+1,i+1
    while j<=r:
        tep[k] = q[j]
        k,j = k+1,j+1
    j = 0
    for i in range(l,r+1):
        q[i] = tep[j]
        j += 1
    return res
n = int(sys.stdin.readline())
while n!=0:
    a = []
    for i in range(n):
        a.append(int(sys.stdin.readline()))
    print(merge_sort(a,0,n-1))
    n = int(sys.stdin.readline())
            
    
    
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