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Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7252 | Accepted: 4478 |
Description
Input
Output
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
Source
#include <cstdio> #include <iostream> #include <sstream> #include <cmath> #include <cstring> #include <cstdlib> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <algorithm> using namespace std; #define ll long long #define _cle(m, a) memset(m, a, sizeof(m)) #define repu(i, a, b) for(int i = a; i < b; i++) #define repd(i, a, b) for(int i = b; i >= a; i--) #define sfi(n) scanf("%d", &n) #define pfi(n) printf("%d\n", n) #define sfi2(n, m) scanf("%d%d", &n, &m) #define pfi2(n, m) printf("%d %d\n", n, m) #define pfi3(a, b, c) printf("%d %d %d\n", a, b, c) #define MAXN 105 const int INF = 0x3f3f3f3f; int a[MAXN]; int dp[MAXN][MAXN]; int main() { int n; while(~sfi(n)) { repu(i, 0, n) sfi(a[i]); _cle(dp, 0x3f); dp[0][0] = a[0]; a[n + 1] = 1; repu(i, 1, n + 1) dp[i][i] = a[i - 1] * a[i] * a[i + 1]; repu(i, 1, n) repu(j, 0, i) dp[i][j] = 0; for(int i = n - 1; i > 0; i--) repu(j, i + 1, n) repu(k, i, j + 1) dp[i][j] = min(dp[i][j], dp[i][k - 1] + a[k] * a[i - 1] * a[j + 1] + dp[k + 1][j]); pfi(dp[1][n - 2]); } return 0; }
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原文地址:http://www.cnblogs.com/sunus/p/4779500.html