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tp<b>R-Tree</b>

  • C++完美演繹 經(jīng)典算法 如 /* 頭文件:my_Include.h */ #include <stdio.h> /* 展開C語言的內(nèi)建函數(shù)指令 */ #define PI 3.141

    C++完美演繹 經(jīng)典算法 如 /* 頭文件:my_Include.h */ #include <stdio.h> /* 展開C語言的內(nèi)建函數(shù)指令 */ #define PI 3.1415926 /* 宏常量,在稍后章節(jié)再詳解 */ #define circle(radius) (PI*radius*radius) /* 宏函數(shù),圓的面積 */ /* 將比較數(shù)值大小的函數(shù)寫在自編include文件內(nèi) */ int show_big_or_small (int a,int b,int c) { int tmp if (a>b) { tmp = a a = b b = tmp } if (b>c) { tmp = b b = c c = tmp } if (a>b) { tmp = a a = b b = tmp } printf("由小至大排序之后的結(jié)果:%d %d %d\n", a, b, c) } 程序執(zhí)行結(jié)果: 由小至大排序之后的結(jié)果:1 2 3 可將內(nèi)建函數(shù)的include文件展開在自編的include文件中 圓圈的面積是=201.0619264

    標(biāo)簽: my_Include include define 3.141

    上傳時(shí)間: 2014-01-17

    上傳用戶:epson850

  • 數(shù)字運(yùn)算

    數(shù)字運(yùn)算,判斷一個(gè)數(shù)是否接近素?cái)?shù) A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value. Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not. Input Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone. Output For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise. Sample Input 10 111 2 110 10 123 6 1000 8 2314 0 Sample Output yes yes no yes no

    標(biāo)簽: 數(shù)字 運(yùn)算

    上傳時(shí)間: 2015-05-21

    上傳用戶:daguda

  • 源代碼用動(dòng)態(tài)規(guī)劃算法計(jì)算序列關(guān)系個(gè)數(shù) 用關(guān)系"<"和"="將3個(gè)數(shù)a

    源代碼\用動(dòng)態(tài)規(guī)劃算法計(jì)算序列關(guān)系個(gè)數(shù) 用關(guān)系"<"和"="將3個(gè)數(shù)a,b,c依次序排列時(shí),有13種不同的序列關(guān)系: a=b=c,a=b<c,a<b=v,a<b<c,a<c<b a=c<b,b<a=c,b<a<c,b<c<a,b=c<a c<a=b,c<a<b,c<b<a 若要將n個(gè)數(shù)依序列,設(shè)計(jì)一個(gè)動(dòng)態(tài)規(guī)劃算法,計(jì)算出有多少種不同的序列關(guān)系, 要求算法只占用O(n),只耗時(shí)O(n*n).

    標(biāo)簽: lt 源代碼 動(dòng)態(tài)規(guī)劃 序列

    上傳時(shí)間: 2013-12-26

    上傳用戶:siguazgb

  • c語言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合

    c語言版的多項(xiàng)式曲線擬合。 用最小二乘法進(jìn)行曲線擬合. 用p-1 次多項(xiàng)式進(jìn)行擬合,p<= 10 x,y 的第0個(gè)域x[0],y[0],沒有用,有效數(shù)據(jù)從x[1],y[1] 開始 nNodeNum,有效數(shù)據(jù)節(jié)點(diǎn)的個(gè)數(shù)。 b,為輸出的多項(xiàng)式系數(shù),b[i] 為b[i-1]次項(xiàng)。b[0],沒有用。 b,有10個(gè)元素ok。

    標(biāo)簽: 多項(xiàng)式 曲線擬合 c語言 最小二乘法

    上傳時(shí)間: 2014-01-12

    上傳用戶:變形金剛

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some tot

    We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.

    標(biāo)簽: represented integers group items

    上傳時(shí)間: 2016-01-17

    上傳用戶:jeffery

  • 用游標(biāo)的方法實(shí)現(xiàn)對(duì)稱差的計(jì)算

    用游標(biāo)的方法實(shí)現(xiàn)對(duì)稱差的計(jì)算,即 (A-B)+(B-A)

    標(biāo)簽: 對(duì)稱 計(jì)算

    上傳時(shí)間: 2016-05-23

    上傳用戶:遠(yuǎn)遠(yuǎn)ssad

  • 詞法分析器 對(duì)輸入一個(gè)函數(shù)

    詞法分析器 對(duì)輸入一個(gè)函數(shù),并對(duì)其分析main() { int a,b a = 10 b = a + 20 }

    標(biāo)簽: 分析器 函數(shù) 輸入

    上傳時(shí)間: 2013-12-20

    上傳用戶:hfmm633

  • 基因算法

    基因算法,用VC++或MATLAB,java等工具設(shè)計(jì)一程序計(jì)算任一個(gè)隨機(jī)產(chǎn)生的DNA基因表達(dá)式的有效長度和值 設(shè)隨機(jī)產(chǎn)生的基因表達(dá)式為: + Q - / b * b a Q b a a b a a b b a a a b

    標(biāo)簽: 基因 算法

    上傳時(shí)間: 2014-01-09

    上傳用戶:aa54

  • 函數(shù)再現(xiàn)機(jī)構(gòu)設(shè)計(jì) 試設(shè)計(jì)一曲柄搖桿機(jī)構(gòu)

    函數(shù)再現(xiàn)機(jī)構(gòu)設(shè)計(jì) 試設(shè)計(jì)一曲柄搖桿機(jī)構(gòu),再現(xiàn)函數(shù) 要求: 輸入構(gòu)件的轉(zhuǎn)角范圍180°,輸出構(gòu)件擺角范圍30°,即: 當(dāng)輸入構(gòu)件從a轉(zhuǎn)至a+90時(shí),輸出構(gòu)件從b轉(zhuǎn)至b+30 當(dāng)輸入構(gòu)件從a+90轉(zhuǎn)至a+180時(shí),輸出構(gòu)件從b+30轉(zhuǎn)至b

    標(biāo)簽: 機(jī)構(gòu) 函數(shù) 曲柄

    上傳時(shí)間: 2013-12-17

    上傳用戶:英雄

  • 這個(gè)連接池是直接從JIVE中取出來的

    這個(gè)連接池是直接從JIVE中取出來的,進(jìn)行了一下修改,使得連接參數(shù)直接在程序中設(shè)定而不是從屬性文件中讀取。 [b]用法:[/b] 先設(shè)定自己的連接參數(shù),在DbConnectionDefaultPool.java文件的loadProperties方法中。注意你也需要設(shè)定連接池的log文件的存放位置。

    標(biāo)簽: JIVE 連接

    上傳時(shí)間: 2016-11-21

    上傳用戶:TF2015

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