题解 | #最小花费爬楼梯#

dp[i]: 走到第i阶需要花费的最少金钱
dp[0]: cost[0];
dp[1]: cost[1];
...
dp[i]: min(dp[i - 1], dp[i - 2]) + cost[i];

const n = parseInt(readline()),
      cost = readline().split(" "),
      dp = [+cost[0], +cost[1]];
for(let i = 2; i < n; i ++){
    dp[i] = Math.min(dp[i - 1], dp[i - 2]) + (+cost[i]);
}
console.log(Math.min(dp[n - 1], dp[n - 2]));