(Large extbf{Description: } large{有n个数的序列,每次可以合并相邻的两个相同的数,得到的一个数原数加一,输出最大能合并多大。(2 leq n leq 248)}\)
(Large extbf{Solution: } large{一道典型的区间DP。我们可以枚举合并区间的长度与起点还有中间断点,然后同时更新答案。值得注意的是,最大值并不一定是 ext{f[1][n]}。\})
(Large extbf{Code: })
#include <cstdio>
#include <algorithm>
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;
const int N = 255;
int n, ans, a, f[N][N];
inline int read() {
char ch = gc();
LL ans = 0;
while (ch > '9' || ch < '0') ch = gc();
while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
return ans;
}
int main() {
n = read();
rep(i, 1, n) a = read(), f[i][i] = a;
rep(i, 2, n)
for (int j = 1; j + i - 1 <= n; ++j) {
int en = i + j - 1;
rep(k, j, en - 1)
if (f[j][k] == f[k + 1][en] && f[j][k] != 0 && f[k + 1][en] != 0) { f[j][en] = max(f[j][en], f[j][k] + 1); ans = max(ans, f[j][k] + 1); }
}
printf("%d
", ans);
return 0;
}