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。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <vector>

 5 #include <cmath>

 6 #include <algorithm>

 7 using namespace std;

 8 #define LL __int64

 9 LL dp[2][51][101];

10 int p[51];

11 int main()

12 {

13     int i,j,k,u,v,n,c,s1,s2,sum;

14     double ans = 0;

15     scanf("%d",&n);

16     sum = 0;

17     for(i = 1; i <= n; i ++)

18     {

19         scanf("%d",&p[i]);

20         sum += p[i];

21     }

22     scanf("%d",&c);

23     if(sum <= c)

24     {

25         printf("%d\n",n);

26         return 0;

27     }

28     for(i = 1; i <= n; i ++)

29     {

30         memset(dp,0,sizeof(dp));

31         dp[0][0][0] = 1;

32         s1 = 0;

33         s2 = 1;

34         for(j = 1; j <= n; j ++)

35         {

36             if(i == j) continue;

37             for(k = 0; k < j; k ++)

38             {

39                 for(u = 0; u <= 100; u ++)

40                     dp[s2][k][u] = dp[s1][k][u];

41             }

42             for(k = 0; k < j; k ++)

43             {

44                 for(u = 0; u < 100; u ++)

45                 {

46                     if(dp[s1][k][u])

47                     {

48                         if(u+p[j] <= 100)

49                             dp[s2][k+1][u+p[j]] += dp[s1][k][u];

50                     }

51                 }

52             }

53             swap(s1,s2);

54         }

55         for(j = 0; j < c; j ++)

56         {

57             if(j + p[i] >= c)

58             {

59                 for(k = 0; k < n; k ++)

60                 {

61                     if(dp[s1][k][j] == 0) continue;

62                     double temp = dp[s1][k][j];

63                     for(u = 1,v = k+1;u <= n-k-1;u ++,v++)

64                     temp *= u*1.0/v;

65                     temp /= n;

66                     if(j + p[i] == c)

67                     ans += temp*(k+1);

68                     else

69                     ans += temp*k;

70                 }

71             }

72         }

73     }

74     printf("%lf\n",ans);

75     return 0;

76 }