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); 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 ",ans); 75 return 0; 76 }