Codeforces 263C. Appleman and Toastman

Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.

解题：贪心，当时乱搞了下，居然对了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 300100;
18 LL d[maxn],sum,ans;
19 int n;
20 int main() {
21     while(~scanf("%d",&n)){
22         for(int i = sum = 0; i < n; i++){
23             cin>>d[i];
24             sum += d[i];
25         }
26         sort(d,d+n);
27         ans = sum;
28         for(int i = 0; i+1 < n; i++){
29             ans += sum;
30             sum -= d[i];
31 
32         }
33         cout<<ans<<endl;
34     }
35     return 0;
36 }

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