bzoj2431: [HAOI2009]逆序对数列(DP)

  f[i][j]前i个数有j个逆序对的数量

  f[i][j]=sigma(f[i-1][j-k]){1<=k<=i}

  维护一个前缀和即可

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio> 
#include<algorithm>
#define ll long long 
using namespace std;
const int maxn=1010,mod=1e4;
int n,k;
int f[maxn][maxn],sum[maxn];
void read(int &k)
{
    int f=1;k=0;char c=getchar();
    while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
    while(c<='9'&&c>='0')k=k*10+c-'0',c=getchar();
    k*=f;
}
inline int MOD(int x){return x>=mod?x-mod:x;}
int main()
{
    read(n);read(k);
    f[0][0]=1;
    for(int i=1;i<=n;i++)
    for(int j=0;j<=min(i*(i-1)>>1,k);j++)
    {
        sum[j]=MOD((j?sum[j-1]:0)+f[i-1][j]);
        f[i][j]=MOD(sum[j]-sum[j-i]+mod);
    }
    printf("%d
",f[n][k]);
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/Sakits/p/7471049.html