POJ 3666 Making the Grade
摘要:
传送门:http://poj.org/problem?id=3666 解题思路: dp[i][j]:代表第i个数以j结尾所花的代价。。那么dp[i][j]=fabs(a[i]-j)+min(dp[i-1][k])k<j; 实现代码: 阅读全文
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