Created
April 11, 2018 15:22
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class Solution { | |
public: | |
double largestSumOfAverages(vector<int>& A, int K) { | |
// const int n = A.size(); | |
// vector<vector<double>> dp(K+1, vector<double>(n+1, 0.0)); | |
// vector<double> sum(n+1, 0.0); | |
// for(int i = 1; i <= n; i++) { | |
// sum[i] = sum[i-1] + A[i-1]; | |
// dp[1][i] = static_cast<double>(sum[i]) / i; | |
// } | |
// for(int k = 2; k <= K; k++) { | |
// for(int i = k; i <= n; i++) { | |
// for(int j = k-1 ; j < i; j++) { | |
// dp[k][i] = max(dp[k][i], dp[k-1][j] + (sum[i] + sum[j])/(i - j)); | |
// } | |
// } | |
// } | |
// return dp[K][n]; | |
const int n = A.size(); | |
vector<vector<double>> dp(K + 1, vector<double>(n + 1, 0.0)); | |
vector<double> sums(n + 1, 0.0); | |
for (int i = 1; i <= n; ++i) { | |
sums[i] = sums[i - 1] + A[i - 1]; | |
dp[1][i] = static_cast<double>(sums[i]) / i; | |
} | |
for (int k = 2; k <= K; ++k) | |
for (int i = k; i <= n; ++i) | |
for (int j = k - 1; j < i; ++j) | |
dp[k][i] = max(dp[k][i], dp[k - 1][j] + (sums[i] - sums[j]) / (i - j)); | |
return dp[K][n]; | |
} | |
}; |
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