bzoj 3029 守卫者的挑战

水题，唯一值得提的是我们不用维护背包剩余空间为O（K）的情况，只要维护最大到n的情况就行了。因为只要剩余空间到了n，无论如何也不会装满。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>

#define md
#define ll long long
#define inf (int) 1e9
#define eps 1e-8
#define N 250
using namespace std;
double f[2][2*N][N];
double p[N];
int a[N];
int main()
{
	int der=210;
	int n,L,K;
	scanf("%d%d%d",&n,&L,&K); f[0][K+der][0]=1;
	for (int i=1;i<=n;i++) { scanf("%lf",&p[i]); p[i]/=100;}
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	for (int i=1;i<=n;i++)
	{
		bool fr,t;
		fr=(i-1)&1; t=fr^1;
		for (int j=der-n;j<=der+n+1;j++)
		  for (int k=0;k<=n;k++)
		    f[t][j][k]=0;
		for (int j=der-n;j<=der+n+1;j++)
		  for (int k=0;k<=i;k++)
		  {
		    f[t][min(j+a[i],der+n+1)][k+1]+=f[fr][j][k]*p[i];
		    f[t][j][k]+=f[fr][j][k]*(1-p[i]);
		  }/*
		for (int j=der-n;j<=der+n+1;j++)
		{
			if (j==der) printf("..."); else printf("   ");
			for (int k=0;k<=i;k++)
			  printf("%.3lf ",f[t][j][k]);
			printf("\n");
		}
		printf("\n");*/
	}
	double ans=0;
	int i=n&1;
	for (int j=der;j<=der+n+1;j++)
	  for (int k=L;k<=n;k++)
	    ans+=f[i][j][k];
	printf("%.6lf\n",ans);
	return 0;
}