逆序数

顾名思义，例如：有序列 2 4 3 1 那么 2 4， 2 1，4 1， 3 1为逆序对，逆序数为4。

#include <stdio.h>#include <assert.h>#include <math.h>#include <string.h>#include <time.h>#include <iostream>#include <sstream>#include <string>#include <vector>#include <map>#include <set>#include <stack>#include <queue>#include <deque>#include <functional>#include <iomanip>#include <algorithm>#include <limits>//#include <ext/hash_map>//using namespace __gnu_cxx;typedef long long LL;typedef unsigned long long ULL;#define RDI(x) scanf("%d",&x)#define RDI_2(x,y) scanf("%d%d",&x,&y)#define RDD(x) scanf("%lf",&x)#define RDD_2(x,y) scanf("%lf%lf",&x,&y)#define RDL(x) scanf("%lld",&x)#define RDL_2(x,y) scanf("%lld%lld",&x,&y)#define RDS(x) scanf("%s",x)#define RDS_2(x,y) scanf("%s%s",x,y)#define PUTI(x) printf("%d",x)#define PUTL(x) printf("%lld",x)#define PUTF(x) printf("%f",x)#define PUTC(x) putchar(x)#define LINE putchar('\n')#define SPACE putchar(' ')#define REP(i,x) for(int i=0;i<int(x);++i)#define REP_1(i,x) for(int i=1;i<=int(x);++i)#define REP_C(i,x) for(i=0;i<int(x);++i)#define REP_E(i,x,y) for(int i=0;i<int(x);++i,y)#define REP_1_E(i,x,y) for(int i=1;i<=int(x);++i,y)#define REP_1_C(i,x) for(i=1;i<=int(x);++i)#define FOR_EDGE(i,x) for(int i=head[x];~i;i=edge[i].next)#define FOR_EDGE2(x,i,y) for(int i=head[x][y];~i;i=edge[i].next)#define FOR(i,a,b) for(int i=int(a);i<=int(b);++i)#define FORD(i,a,b) for(int i=int(a);i>=int(b);--i)#define DWN(i,x) for(int i=int(x-1);i>=0;--i)#define DWN_1(i,x) for(int i=int(x);i>0;--i)#define FIL(x,v) memset(x,v,sizeof(x))#define CLR(x) memset(x,0,sizeof(x))#define ALL(x) x.begin(),x.end()#define p_queue priority_queue#define pb push_back#define pf push_front#define popb pop_back#define popf pop_front#define fst first#define snd second#define x first#define y second#define mp(x,y) make_pair(x,y)//#define check(x) printf("%s=%d\n",#x,x)using namespace std;template<typename T>T gcd(T a, T b){ return b == 0 ? a : gcd(b, a%b); }template<typename T>T lcm(T a, T b){ return a / gcd(a, b)*b; }template<typename T>T ext_gcd(T a, T b, T& x, T& y){    if (b == 0)    {        x = 1, y = 0;        return a;    }    int d = ext_gcd(b, a%b, y, x);    y -= x*(a / b);    return d;}template<typename T>T quick_pow(T a, T b){    if (b == 0) return 1;    T p = quick_pow(a, b >> 1);    p = p*p;    if (b & 1) p =p*a;    return p;}double combination(int n, int r){    if (r<n - r) return combination(n, n - r);    double s = 1.0;    REP_1(i, n - r) s *= (r + i), s /= i;    return s;}template<typename T>inline void read(T& v){    T ret = 0, sgnv = 1;    char p;    while (!isdigit(p = getchar())) if (p == '-') sgnv = -1;    do{        ret = (ret << 3) + (ret << 1) + p - '0';    } while (isdigit(p = getchar()));    v = sgnv*ret;}const double eps=1e-8;inline int sgn(double x){ return (x>-eps)-(x<eps); }const int mod=1000000007;//const int INF = numeric_limits<int>::max();const int INF = 0x7f7f7f7f;//const int INFCELL = 0x7f;#define MAX 1000+5#define lucknum 19960308#define ONLINE_JUDGEvoid merge(int *a, int *t, int leftPos, int rightPos, int rightEnd, int &ans){    int leftEnd = rightPos - 1;    int elementsNum = rightEnd - leftPos + 1;    int tmpPos = leftPos;    while(leftPos <= leftEnd && rightPos <= rightEnd)    {        if(a[rightPos] < a[leftPos])    //此时产生逆序数        {            ans += leftEnd - leftPos + 1;            t[tmpPos++] = a[rightPos++];        }        else            t[tmpPos++] = a[leftPos++];    }    while(leftPos <= leftEnd)   t[tmpPos++] = a[leftPos++];    while(rightPos <= rightEnd) t[tmpPos++] = a[rightPos++];    for(int i = 0; i < elementsNum; ++i, --rightEnd)        a[rightEnd] = t[rightEnd];}void mergeSort(int *a, int *t, int left, int right, int &ans){    if(left < right)    {        int center = (left + right) >> 1;        mergeSort(a, t, left, center, ans);        mergeSort(a, t, center + 1, right, ans);        merge(a, t, left, center + 1, right, ans);    }}int main(){#ifndef ONLINE_JUDGE    freopen("input.in","r",stdin);#endif    int a[MAX], t[MAX], n, ans = 0;    RDI(n);    REP(i, n)   RDI(a[i]);    mergeSort(a, t, 0, n - 1, ans);    PUTI(ans);    return 0;}