56 lines
1.5 KiB
C++
56 lines
1.5 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 55;
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typedef __int128 LL;
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const LL INF = 1e38;
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LL read() {
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LL x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = x * 10 + ch - '0';
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ch = getchar();
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}
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return x * f;
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}
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void write(LL x) {
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if (x < 0) putchar('-'), x = -x;
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if (x > 9) write(x / 10);
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putchar(x % 10 + '0');
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}
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LL f[N][N];
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LL w[N];
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int n;
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int main() {
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scanf("%d", &n);
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for (int i = 1; i <= n; i++) w[i] = read();
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// 初始化
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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// 两个点之间最小差距是2，比如[1,3]这里面就有三角形，需要求解，如果小于2，比如[1,2]
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// 无法构成三角形，就不需要求解，无法构成的情况贡献值是0，也是递推的起点，否则，全都是INF
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// 就无法推起来了
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j - i < 2 ? f[i][j] = 0 : f[i][j] = INF;
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for (int len = 3; len <= n; len++) { // 区间范围最少是3个，否则无法构建三角形
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for (int l = 1; l + len - 1 <= n; l++) {
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int r = l + len - 1;
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for (int k = l + 1; k < r; k++) // l<k<r才能构建三角形
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f[l][r] = min(f[l][r], f[l][k] + f[k][r] + w[l] * w[k] * w[r]);
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}
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}
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// 输出
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write(f[1][n]);
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return 0;
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}
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