There is a row of n coins whose values are some positive integers c?, c?,...,cn, not necessarily distinct. The goal is to pick up the maximum amount of money subject to the constraint that no two coins adjacent in the initial row can be picked up.
输入
Two lines, the first line is n (0< n <=10000), and the second line is value of coin(0< value <= 2^32).
输出
the maximum amount of money.
样例输入
6 5 1 2 10 6 2
样例输出
17
思路：估计这老师才讲了dp 算法吧,这组题几乎全是dp（也是醉了）
不能够相邻那么就考虑当前位的前两个数对应的最好状态+当前数，和当前数的前一个数的状态比较，这里的边界dp[0]=0,dp[1]=x[1](x数据元素数组从1开始存贮)，
那么可以得到如下dp方程 dp[i] = max(dp[i - 1], dp[i - 2] + x[i]);
#include <iostream> using namespace std; int n, dp[10001], i, x[10001]; #define max(a,b) a>b?a:b int main() { cin >> n; for (i = 1; i <= n; i++) cin >> x[i]; dp[1] = x[1]; for (i = 2; i <= n; i++) dp[i] = max(dp[i - 1], dp[i - 2] + x[i]); cout << dp[n] << "\r\n"; return 0; }