#1141: Two-point and merge-sorted inverse pairs (merge sort)

#1141: Two-point and merge-sort in reverse order
Time limit: 10000ms
Single point time limit: 1000ms
Memory Limit: 256MB

Describe

We knew Nettle was playing "Battleship これ" on the previous, last, and upper back. After a bitter struggle, nettle again got a lot of boats.
This day nettle is checking its fleet list:
[List.png]
As we can see, the ship's default sort is the rank parameter. But in fact a ship's fire value and the level of the relationship is not big, so there will be a ship than B ship grade, but a ship fire is lower than the case of B ship. For example, the 77-stage dragon to two firepower is less than the 55-level of the XI-Li two.
Now that the nettle will give all the ship's firepower in order of rank, please calculate how many pairs of ships meet the above mentioned situation.

Tip: High firepower is fierce!
Input

Line 1th: an integer n. N indicates the number of ships, 1≤n≤100,000
Line 2nd: n integers, the number of I indicates the fire value of the ship with low rank a[i],1≤a[i]≤2^31-1.
Output

Line 1th: An integer that indicates how many pairs of ships meet "a ship is higher than B ship, but a ship's firepower is lower than ship B."
Sample input

10
1559614248 709366232 500801802 128741032 1669935692 1993231896 369000208 381379206 962247088 237855491

Sample output

27

#include <cstdio> #include <algorithm>//#include <bits/stdc++.h>using namespace std;template<    Class T>inline T Read (t&x) {char C;    while ((C=getchar ()) <=32) if (c==eof) return 0;    BOOL Ok=false;    if (c== '-') Ok=true,c=getchar ();    for (x=0; c>32; C=getchar ()) x=x*10+c-' 0 ';    if (OK) x=-x; return 1;} Template<class t> inline T read_ (t&x,t&y) {return read (x) &&read (y);} Template<class t> inline T read__ (t&x,t&y,t&z) {return read (x) &&read (y) &&read (z);}    Template<class t> inline void Write (T x) {if (x<0) Putchar ('-'), x=-x;    if (x<10) putchar (x+ ' 0 '); else write (X/10), Putchar (x%10+ ' 0 ');}    Template<class t>inline void Writeln (T x) {write (x); Putchar (' \ n ');} -------ZCC IO template------const int MAXN=100011;CONST double inf=999999999; #define Lson (rt<<1), L,m#define Rson (rt<<1|1), M+1,r#define M ((l+r) >>1) #define for (i,t,n) for (int i= (t);i< (n); i++) TypEdef Long Long ll;typedef double db;typedef pair<int,int> P; #define BUG printf ("---\ n"); #define MOD 10007LL ans=0;/        /int is not an existing void Merge_sort (int *a,int x,int y,int *t) {if (y-x>1) {int p=x,m= (x+y)/2;        int i=x;        int q=m;        Merge_sort (a,x,m,t);        Merge_sort (a,m,y,t); while (p<m| | Q<y) {if (q>=y| | (P<m&&a[p]<=a[q]))            T[i++]=a[p++];        else {t[i++]=a[q++];ans+=m-p;}    } for (int i=x;i<y;i++) a[i]=t[i]; }} int A[MAXN];    int B[maxn];int Main () {//#ifndef Online_judge//Freopen ("In.txt", "R", stdin);    #endif//Online_judge int n,m,i,j,k,t;        while (read (n)) {for (i,0,n) {read (a[i]);        } ans=0;        Merge_sort (A,0,N,B);    Writeln (ANS); } return 0;}

#1141: Two-point and merge-sorted inverse pairs (merge sort)