BZOJ3427 Poi2013 Bytecomputer

可以YY一下嘛= =

最后一定是-1, -1, ..., -1, 0, 0, ... 0, 1, 1, ..., 1的一个数列

于是f[i][j]表示到了第i个数，新数列的第j项为-1 or 0 or 1的的最小代价

然后就没有然后了

 /**************************************************************
     Problem: 3427
     User: rausen
     Language: C++
     Result: Accepted
     Time:808 ms
     Memory:16432 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int N = ;
 const int inf = 1e9;

 int a[N], f[N][];
 int n, ans = inf;

 inline int read() {
     int x = , sgn = ;
     char ch = getchar();
     while (ch < '' || '' < ch) {
         if (ch == '-') sgn = -;
         ch = getchar();
     }
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return sgn * x;
 }

 int main() {
     int i, j, k;
     int t, to;
     n = read();
     for (i = ; i <= n; ++i)
         a[i] = read();
     memset(f, , sizeof(f));
     f[][a[] + ] = ;
     for (i = ; i < n; ++i)
         for (j = ; j <= ; ++j) if (f[i][j] < inf)
             for (t = j - , k = ; k <= ; ++k) {
                 to = a[i + ] + k * t;
                 if (to >= - && to <=  && to >= t)
                     f[i + ][to + ] = min(f[i + ][to + ], f[i][j] + k);
             }
     for (i = ; i <= ; ++i)
         ans = min(ans, f[n][i]);
     if (ans == inf) puts("BRAK");
     else printf("%d\n", ans);
     return ;
 }