Codeforces Round #160 (Div. 2) D. Maxim and Restaurant(DP)

题目链接

想了挺久，枚举每一件物品，当做关键物品，假设再加这一件物品，就>=c了，把剩下的物品背一下包，dp[i][j]表示i个物品可以组成重量j的个数。

这样就可以知道前面放i件，后边肯定放n-i-1件，乱搞搞，算double，边乘边算保证不要越界。最后注意，LL和sum <= c的时候情况。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define LL __int64
 LL dp[][][];
 int p[];
 int main()
 {
     int i,j,k,u,v,n,c,s1,s2,sum;
     double ans = ;
     scanf("%d",&n);
     sum = ;
     for(i = ; i <= n; i ++)
     {
         scanf("%d",&p[i]);
         sum += p[i];
     }
     scanf("%d",&c);
     if(sum <= c)
     {
         printf("%d\n",n);
         return ;
     }
     for(i = ; i <= n; i ++)
     {
         memset(dp,,sizeof(dp));
         dp[][][] = ;
         s1 = ;
         s2 = ;
         for(j = ; j <= n; j ++)
         {
             if(i == j) continue;
             for(k = ; k < j; k ++)
             {
                 for(u = ; u <= ; u ++)
                     dp[s2][k][u] = dp[s1][k][u];
             }
             for(k = ; k < j; k ++)
             {
                 for(u = ; u < ; u ++)
                 {
                     if(dp[s1][k][u])
                     {
                         if(u+p[j] <= )
                             dp[s2][k+][u+p[j]] += dp[s1][k][u];
                     }
                 }
             }
             swap(s1,s2);
         }
         for(j = ; j < c; j ++)
         {
             if(j + p[i] >= c)
             {
                 for(k = ; k < n; k ++)
                 {
                     if(dp[s1][k][j] == ) continue;
                     double temp = dp[s1][k][j];
                     for(u = ,v = k+;u <= n-k-;u ++,v++)
                     temp *= u*1.0/v;
                     temp /= n;
                     if(j + p[i] == c)
                     ans += temp*(k+);
                     else
                     ans += temp*k;
                 }
             }
         }
     }
     printf("%lf\n",ans);
     return ;
 }