PO65最后的士兵
作者:互联网
思路:动态规划
#include <vector>
#include <numeric>
using namespace std;
class Solution
{
public:
int lastStoneWeightII(vector<int> &nums)
{
int sum = accumulate(nums.begin(), nums.end(), 0);
int n = nums.size(), m = sum / 2;
vector<vector<int> > dp(n + 1, vector<int>(m + 1));
dp[0][0] = true;
for (int i = 0; i < n; ++i)
{
for (int j = 0; j <= m; ++j)
{
if (j < nums[i])
{
dp[i + 1][j] = dp[i][j];
}
else
{
dp[i + 1][j] = dp[i][j] || dp[i][j - nums[i]];
}
}
}
for (int j = m;; --j)
{
if (dp[n][j])
{
return sum - 2 * j;
}
}
}
};
int main()
{
Solution solution;
vector<int>nums;
int n, temp;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> temp;
nums.push_back(temp);
}
cout << solution.lastStoneWeightII(nums) << endl;
return 0;
}
-[1]最后一块石头的重量 II
标签:temp,nums,int,sum,最后,PO65,vector,士兵,include 来源: https://www.cnblogs.com/aoke2002/p/15270535.html