- 简单动态规划(一)
第一题
- 2025-6-7 14:31:21 @
#include <bits/stdc++.h>
using namespace std;
int a[41],n,dp[50];
int main(){
scanf("%d",&n);
for(int i = 1;i<=n;i++)scanf("%d",&a[i]);
for(int i = 1;i<=n;i++)dp[i] = min(dp[i-1]+a[i],dp[i-2]+a[i]);
printf("%d",min(dp[n],dp[n-1]));
return 0;
}
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