【HDU 4911 Inversion】 树状数组

HDU 4911
题意就是叫你求逆序数 然后可以交换k 次 问你最少剩多少逆序数
其实就是求个逆序数 求个 max(0,逆序数-k)

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 100025;
int arr[MAX_N],b[MAX_N],C[MAX_N<<1];
void add(int x,int v)
{
    for(;x<MAX_N;x+=x&(-x))
        C[x] += v;
}
int getsum(int x)
{
    int res = 0;
    for(;x;x-=x&(-x))
        res+=C[x];
    return res;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,k;
    ll ans ;
    while(scanf("%d%d",&n,&k)==2)
    {
        ans = 0;
        memset(C,0,sizeof(C));
        for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),b[i] = arr[i];
        sort(b+1,b+1+n);
        int sz = unique(b+1,b+1+n)-b-1;
        for(int i = 1;i<=n;++i) arr[i] = lower_bound(b+1,b+1+sz,arr[i])-b+1;
        for(int i = 1;i<=n;++i)
        {
            ans+=i-1-getsum(arr[i]);
            add(arr[i],1);
        }
        printf("%lld\n",max(0ll,ans-k));
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

标签：HDU,const,int,ll,Inversion,4911,ans,include,mod
来源： https://blog.csdn.net/heucodesong/article/details/89763942