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[USACO2005][poj2229]Sumsets(递推)
http://poj.org/problem?id=2229 分析: 显然的递推 若n为奇数,那么肯定是在n-1的基础上前面每个数+1,即f[n]=f[n-1] 若n为偶数 当第一位数字是1的时候,等同于f[n-1] 当第一位数字不是1的时候,因为都是2的倍数,可以提出一个2来,即与f[n/2]相同 综上,n为偶数时候,f[n]=f[n-1]+f[n/2]
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原文地址:https://www.cnblogs.com/wmrv587/p/3588832.html
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