[luoguP2513] [HAOI2009]逆序对数列（DP）

传送门

f[i][j]表示前i个数，逆序对数为j的答案

则DP方程为：

f[1][0] = 1;
    for(i = 2; i <= n; i++)
        for(j = 0; j <= m; j++)
            for(k = j; k < j + i; k++)
                f[i][k] = (f[i][k] + f[i - 1][j]) % p;

但是会超时

所以搞个前缀和优化一下

#include <cstdio>
#include <iostream>
#define N 2001
#define p 10000

int n, m;
int f[N][N], sum[N][N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

int main()
{
	int i, j, k;
	n = read();
	m = read();
	f[1][0] = 1;
	for(i = 0; i <= m; i++) sum[1][i] = 1;
	for(i = 2; i <= n; i++)
		for(j = 0; j <= m; j++)
		{
			if(j - i + 1 > 0)
				f[i][j] = (f[i][j] + ((sum[i - 1][j] - sum[i - 1][j - i]) % p + p) % p) % p;
			else
				f[i][j] = (f[i][j] + sum[i - 1][j]) % p;
			sum[i][j] = (sum[i][j - 1] + f[i][j]) % p;
		}
	printf("%d
", f[n][m]);
	return 0;
}