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=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),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<<1;return 0;}
    for (int j=0;j<=k;j++) f[0][j]=1;
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<=k;j++)
        inc(f[i][j],((j-i>=0?f[i-1][j]-f[i-1][j-i]:f[i-1][j])+P)%P);
        for (int j=1;j<=k;j++)
        inc(f[i][j],f[i][j-1]);
    }
    cout<<(f[n][k]-f[n][k-1]+P)%P;
    return 0;
}