BZOJ2431 HAOI2009逆序对数列（动态规划）

　　对于排列计数问题一般把数按一个特定的顺序加入排列。这个题做法比较显然，考虑将数从小到大加入排列即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
#define P 10000
int n,k,f[N][N];
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj2431.in","r",stdin);
    freopen("bzoj2431.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),k=read();if (!k) {cout<<;return ;}
    for (int j=;j<=k;j++) f[][j]=;
    for (int i=;i<=n;i++)
    {
        for (int j=;j<=k;j++)
        inc(f[i][j],((j-i>=?f[i-][j]-f[i-][j-i]:f[i-][j])+P)%P);
        for (int j=;j<=k;j++)
        inc(f[i][j],f[i][j-]);
    }
    cout<<(f[n][k]-f[n][k-]+P)%P;
    return ;
}