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pingdingchuang-fuzhi-xiuzheng

于 2012-07-23 发布 文件大小:6KB
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代码说明:

  不同的窗函数,都对应不同的幅值修正系数;针对复制修正中最重要的平顶窗,给出修正的方法。(Different window functions, corresponding to the amplitude correction factor copy the most important amendment of the flat-top windows, give a correction method.)

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  • knapsack
    实现背包的最有装载,可以实现背包的最有价值(The realization of the most loaded backpack, the backpack can be the most valuable)
    2008-12-12 10:39:41下载
    积分:1
  • 0
    说明:  给定n种物品和一个背包,物品i的重量是Wi,价值是vi,被包容量是C,应该如何选择装入被包的物品,使得装入背包中物品的总价值最大?对于每种物品i,只有两种选择,装入或者不装入. 输入:第一行两个正整数n(物品数)和C(容量) 第二行n个正整数n,表示物品价值 第三行n个正整数n,表示物品重量 输出:第一行表示总价值 第二行n个数(0/1),0表示物品不装入,1表示装入 例如, 输入: 5 10 6 3 5 4 6 2 2 6 5 4 输出 15 1 1 0 0 1 (Species of a given n items and a backpack, the weight of item i is Wi, the value of vi, was packet capacity is C, should be how to choose a package of items to be loaded, the backpack load in the largest total value of goods? For each item i, there are only two options, loaded or not loaded. Input: the first line of the two positive integer n (the number of items) and C (capacity) of the second line of n positive integer n, that the value of the third line of n positive integer n, that the weight of output: the total value of the first line of the second line, said n number of (0/1), 0 that the items do not load, that load 1 for example, enter: 5,106,354,622 6 5 4 output 15 1 1 0 0 1)
    2009-07-01 23:07:01下载
    积分:1
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    2011-06-20 16:34:15下载
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    2011-09-28 11:06:53下载
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