BZOJ 2431: [HAOI2009]逆序对数列

设 \(dp_{i,j}\) 为长度为 \(i\) 的排列，逆序对数有 \(j\) 个的方案数
那么 \(dp_{i,j}=\sum\limits_{k=\max\{0,j-(i-1)\}}^j dp_{i-1, k}\)
考虑每个新加入的数插入的位置对逆序对数的影响即可得到上述转移方程
通过前缀和优化能实现 \(O(1)\) 转移
总复杂度与状态数同阶

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define SZ(x) ((int)(x).size())
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=((b)-1);i>=(a);i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 10000,INF=0x3f3f3f3f;
const ll inf=0x3f3f3f3f3f3f3f3f;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2, int mod = MOD) {int ans = 1; for (; b; a = 1LL * a * a % mod, b >>= 1)if (b & 1)ans = 1LL * ans * a % mod; return ans % mod;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 1e3 + 7;
int dp[N][N], sum[N][N];
int n, k;

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	n = _(), k = _();
	dp[0][0] = 1;
	sum[0][0] = 1;
	rep (i, 0, k + 1) sum[0][i] = 1;
	rep (i, 1, n + 1) {
		rep (j, 0, k + 1) {
			M(dp[i][j] += sum[i - 1][j]);
			int l = std::max(0, j - (i - 1));
			if (l > 0) M(dp[i][j] -= sum[i - 1][l - 1]);
			M(sum[i][j] += dp[i][j]);
			if (j) M(sum[i][j] += sum[i][j - 1]);
		}
	}
	printf("%d\n", dp[n][k]);
#ifdef LOCAL
	printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
	return 0;
}