题解 | #最小花费爬楼梯#

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <map>
#include <queue>
#include <cmath>
using namespace std;

int cost[100005];
int dp[100005]; //dp[i] 到达第i层时，所花费的最小值

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        for (int i = 0; i < n; i++) {
            scanf("%d", &cost[i]);
            dp[i] = cost[i];
        }
        dp[0] = 0;
        dp[1] = 0; //从0层和1层都可以开始，花费为0
	  
        for (int i = 2; i <= n; i++) {
            dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }
  
        printf("%d\n", dp[n]);
    }
}