BZOJ 2431 [HAOI2009]逆序对数列 (dp)

题意

问长度为n的1~n的排列，且逆序对为k的方案数有多少
(n,kleq 1000)

思路

假设前(1)~(i)已经排列好,此时逆序对为(k)，那么我们来讨论插入(i+1)时候的状态
(i+1)根据插入位置的不同，可以产生(0)到(i)个逆序对
根据这个特点我们设(dp[i][j])为前(i)个数，逆序对为(j)的排列的方案数
(dp[i][j]=f[i-1][j-i+1]+f[i-1][j-i+2]+cdots +f[i-1][j])

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
//#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
 
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 10000;
const int maxn = 5e5+100;
const int maxm = 2e6+100;
const int inf = 0x7f3f3f3f;
//const db pi = acos(-1.0);
const ull base = 201326611;

int f[1111][1111];
int sum[1111];
int n,k;
int main() {
    scanf("%d %d", &n, &k);
    f[1][0]=1;
    sum[0]=1;
    for(int j = 1; j <= 1000; j++){
        sum[j]=sum[j-1];
        sum[j]%=mod;
    }
    for(int i = 2; i <= n; i++){
        f[i][0]=1;
        for(int j = 1; j <= 1000; j++){
            int g = j-i;
            if(j-i<0)g=0;
            else g=sum[g];
            f[i][j]=sum[j]-g+mod;
            f[i][j]%=mod;
        }
        sum[0]=1;
        for(int j = 1; j <= 1000; j++){
            sum[j]=sum[j-1]+f[i][j];
            sum[j]%=mod;
        }
    }

    printf("%d",f[n][k]);
    return 0;
}