1 /*
2 s[i]表示前i个的和
3 d[i][j]表示第i个到第j个直接花费的最小代价
4
5 d[i][j] = min(d[i][k] + d[k][j])(k from i to j)
6
7 */
8 #include<cstdio>
9 #define inf 1<<30
10 int s[205],d[205][205];
11 int main()
12 {
13 int n,i,j,k,x,t,temp;
14 while(~scanf("%d",&n))
15 {
16 x = 0;
17 for(i=1; i<=n; ++i)
18 {
19 d[i-1][i] = x;
20 scanf("%d",&x);
21 d[i-1][i] += x;
22 s[i] = x + s[i-1];
23 }
24 for(i=2; i<n; ++i)//控制区间元素个数为i+1个
25 {
26 for(j=i+1; j<=n; ++j)//区间的最后一个元素
27 {
28 t = inf;
29 for(k=j-i; k<j; ++k)//区间分割点处(j-i <= k < j)
30 {
31 temp = d[j-i][k] + d[k+1][j] + s[j] - s[j-i-1];
32 if(temp < t)//找到[j-i,j]的最小价值。
33 t = temp;
34 }
35 d[j-i][j] = t;
36 }
37 }
38 printf("%d\n",d[1][n]);
39 }
40 return 0;
41 }
42
43 //最优解，暂时没看懂。。。。
44 #include <cstdio>
45 const int N=210;
46 int n,t,stone[N],ans;
47 void combine(int k)
48 {
49 int tmp=stone[k]+stone[k-1];
50 ans+=tmp;
51 for(int i=k; i<t-1; i++)
52 stone[i]=stone[i+1];
53 t--;
54 int j;
55 for(j=k-1; j>0 && stone[j-1]<tmp; j--)
56 stone[j] = stone[j-1];
57 stone[j] = tmp;
58 while(j >= 2 && stone[j] >= stone[j-2])
59 {
60 int d = t - j;
61 combine(j-1);
62 j = t-d;
63 }
64 }
65 int main()
66 {
67 while(~scanf("%d",&n))
68 {
69 for(int i=0; i<n; i++) scanf("%d",&stone[i]);
70 t=1,ans=0;
71 for(int i=1;i<n;i++)
72 {
73 stone[t++]=stone[i];
74 while(t >=3 && stone[t-3]<=stone[t-1])
75 combine(t-2);
76 }
77 while(t > 1) combine(t-1);
78 printf("%d\n" , ans);
79 }
80 return 0;
81 }
