bzoj2431[HAOI2009]逆序对数列

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我们用f[i][j]来表示前i个数,有j对逆序对的方案数,so考虑插入,因为i比1到i-1里面的任何一个数都要大,所以插入的时候后面有多少个数,就会多出多少逆序对,所以前面的符合条件的都可以转移,f[i][j]=sigma(f[i-1][j-k]),k属于0到i-1,然而前缀和神烦.

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define LL long long#define fo(i,a,b) for(int i=a;i<=b;i++)using namespace std;#define N 1005#define inf 10000int f[N][N],s[N][N];int n,m;int main(){scanf("%d%d",&n,&m);memset(f,0,sizeof(f));memset(s,0,sizeof(s));f[1][0]=1;fo(i,0,m)s[1][i]=1;fo(i,2,n){fo(j,0,m){f[i][j]=s[i-1][j];f[i][j]%=inf;if(j-i>=0){f[i][j]-=s[i-1][j-i];f[i][j]+=inf;f[i][j]%=inf;}if(j)s[i][j]=s[i][j-1];s[i][j]+=f[i][j];s[i][j]%=inf;//s[i][j]=f[i][j];else s[i][j]=s[i][j-1]+f[i][j];}}cout<<f[n][m]<<endl;return 0;}