1523. K-inversions(K逆序对）

1523

这题应该说有一些DP的思想吧 dp[i][j]表示以i为结尾第j个数的个数 k单调下降 直接求的话肯定超时 然后用树状数组来进行维护

求k-1次树状数组

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;
 #define lowbit(x) x&(-x)
 #define N 20010
 #define LL long long
 #define mod 1000000000
 int dp[N][],n,a[N],po[N],sum[N][];
 void add(int x,int k,int d)
 {
     while(x)
     {
         dp[x][k] = (d+dp[x][k])%mod;
         x-=lowbit(x);
     }
 }
 LL getsum(int x,int k)
 {
     LL s=;
     while(x<=n)
     {
         s=(s+dp[x][k])%mod;
         x+=lowbit(x);
     }
     return s;
 }
 int main()
 {
     int i,j,k;
     scanf("%d%d",&n,&k);
     for(i = ;i <= n ; i++)
     {
         scanf("%d",&a[i]);
         sum[a[i]][] = ;
     }
     LL s = ;
     for(i = ; i <= k ; i++)
     {
         for(j = i-; j <= n ;j++)
         {
             add(a[j],i,sum[a[j]][i-]);
             sum[a[j]][i] = getsum(a[j]+,i);
         }
     }
     for(i =  ; i <= n ; i++)
     s = (s+sum[a[i]][k])%mod;
     printf("%lld\n",s);
     return ;
 }