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#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
const int N = 60, M = 50;
int n;
int w[N];
LL f[N][N][M];
void add(LL a[], LL b[])
{
LL c[M];
memset(c, 0, sizeof c);
LL t = 0;
for (int i = 0; i < M; i ++) {
t += a[i] + b[i];
c[i] = t % 10;
t /= 10;
}
memcpy(a, c, sizeof c);
}
void mul(LL a[], LL b)
{
LL c[M];
memset(c, 0, sizeof c);
LL t = 0;
for (int i = 0; i < M; i ++) {
t += a[i] * b;
c[i] = t % 10;
t /= 10;
}
memcpy(a, c, sizeof c);
}
int cmp(LL a[], LL b[])
{
for (int i = M - 1; i >= 0; i --) {
if (a[i] > b[i])
return 1;
else if (a[i] < b[i])
return -1;
}
return 0;
}
void print(LL a[])
{
int k = M - 1;
while (k && !a[k])
k --;
while (k >= 0)
cout << a[k --];
cout << endl;
}
int main()
{
cin >> n;
for (int i = 1; i <= n; i ++)
cin >> w[i];
LL tmp[M];
for (int len = 3; len <= n; len ++) {
for (int l = 1; l + len - 1 <= n; l ++) {
int r = l + len - 1;
f[l][r][M - 1] = 1;
for (int k = l + 1; k < r; k ++) {
int t = w[l];
for (int i = 0; i < M; i ++) {
tmp[i] = t % 10;
t /= 10;
}
mul(tmp, w[k]);
mul(tmp, w[r]);
add(tmp, f[l][k]);
add(tmp, f[k][r]);
if (cmp(tmp, f[l][r]) < 0)
memcpy(f[l][r], tmp, sizeof tmp);
}
}
}
print(f[1][n]);
return 0;
}
- 状态表示
/(f[l][r]/) 表示将顶点 /([l,r]/) 划分成三角形的所有方案的最大值 - 状态计算
用 /(k/) 表示 /((l,r)/) 中的一点
/(f[l][r] = min(f[l][k] + f[k][r] + w[l] /cdot w[k] /cdot w[r])/) - 高精度写法
原创文章,作者:,如若转载,请注明出处:https://blog.ytso.com/272182.html