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Search - INT 1 - List
[
Mathimatics-Numerical algorithms
]
int
DL : 0
一个高精度库,重载了一些操作符
Date
: 2008-10-13
Size
: 1.41kb
User
:
xp
[
Mathimatics-Numerical algorithms
]
super-int-operation
DL : 0
此算法实现超长整数(计算机所不能表示的整数,如长度为100位、1000位)的加、减、乘、除运算!
Date
: 2008-10-13
Size
: 1.58kb
User
:
孟繁荣
[
Mathimatics-Numerical algorithms
]
遗传算法SGA
DL : 0
包含一个cpp文件,含有main()函数! 主要函数如下 void InitData(); int Flip(float probabiliby); float Random(); //generate a pseudorandom integer from 0 to 1 void ResetRandom(); //reset pseudorandom integer array fRand float ObjFunc(float vx); float DeCode(unsigned * pChrom); void StatPop(POP * pop); void InitPop(); //initialize population void InitReport(); //initial info report unsigned Select(); int Mutation(unsigned chromval); int CrossOver(unsigned * parent1,unsigned * parent2,int popidx); void UpdateGen(); void Report(int gen);
Date
: 2009-05-23
Size
: 525.05kb
User
:
arkzhu
[
Mathimatics-Numerical algorithms
]
约瑟夫
DL : 0
#include void main() { int i,j,k,a[31],n=0; for(i=0;i<=30;i++) a[i]=0; k=1; for(i=1;i30) k=1; } a[k++]=1; if(k>30) k=1; //n++; } printf("%d\n",k); for(i=1;i<=30;i++) printf("%d ",a[i]); printf("(1为非*,0为*)"); }
Date
: 2009-06-16
Size
: 6.26kb
User
:
man001
[
Mathimatics-Numerical algorithms
]
mt19937int.c
DL : 0
伪随机数生成算法,很优秀,32位int-pseudo-random number generation algorithms, excellent, 32 int
Date
: 2026-01-06
Size
: 2kb
User
:
张清之
[
Mathimatics-Numerical algorithms
]
lcs(continue)
DL : 0
LCS,即最常公共子序列的的C语言解法。prepare_for_backdate(char,char,int,int)函数是为后面的回溯法求得最长公共子序列做准备,并可得到子序列长度。lcs(char,int,int)函数是输出子序列的。并用到了第一个函数的结果。因为要得到最终的子序列,要知道那些地方是可输出的位置,因此构造数组b[][],当为1时表明当前位置匹配,可输出,为2时需要往上回溯,为3时需要往左回溯,直到找到下一个为1的位置。而c[][]数组是保存找子序列过程中匹配位数。-LCS, that is the most common public sequence of C language method. prepare_for_backdate (char, char, int, int) function is obtained by back behind the longest sequence to prepare the public and may be sub-sequence length. lcs (char, int, int) function is the output sub-sequences. And use the first results of a function. Because to obtain the final sequence, it is necessary to know where is the location of the output, so the array structure b [][], as 1:00 shows that match the current location can be output, 2:00 needs to back up for the 3:00 need to go back to the left until you find the next 1 position. And c [] [] array is to preserve the process of looking for sequence matching the median.
Date
: 2026-01-06
Size
: 1.01mb
User
:
向冬冬
[
Mathimatics-Numerical algorithms
]
super-int-operation
DL : 0
此算法实现超长整数(计算机所不能表示的整数,如长度为100位、1000位)的加、减、乘、除运算!-Long Integer realize this algorithm (the computer should not express an integer, such as a length of 100, 1000) of the add, subtract, multiply, with the exception of computing!
Date
: 2026-01-06
Size
: 1kb
User
:
孟繁荣
[
Mathimatics-Numerical algorithms
]
shiyan12
DL : 0
编制具有如下原型的函数prime,用来判断整数n是否为素数:bool prime(int n) 而后编制主函数,任意输入一个大于4的偶数d,找出满足d=d1+d2的所有数对,其中要求d1与d2均为素数(通过调用prime来判断素数)。如偶数18可以分解为11+7以及13+5;而偶数80可以分解为:43+37、61+19、67+13、73+7。 提示:i与d-i的和恰为偶数d,而且只有当i与d-i均为奇数时才有可能成为所求的“数对”。 -Prepared with the following function prototype prime, used to determine whether an integer n prime numbers: bool prime (int n) before the preparation of the main function, indiscriminate importation of a greater than 4, even d, to find out to satisfy d = d1+ D2 all the number of , which calls for d1 and d2 are prime numbers (by calling the prime to determine prime numbers). If even 18 can be decomposed into 11+ 7 and 13+ 5 and even 80 can be decomposed into: 43+ 37,612 B! 19,672 B! 13,732 B! 7. Tip: i with di and exactly even d, and only when i and di are odd when they may become the order of a few of.
Date
: 2026-01-06
Size
: 883kb
User
:
[
Mathimatics-Numerical algorithms
]
VCVB
DL : 0
本书是所随带的VC++和BC++数值分析类库光盘的使用手册。此VC++和BC++数值分析类库涵盖了数值分析领域中大部分常见算法,还包括线性和非线性最优化问题的多种算法以及概率统计中的一些基本算法。此类库中将矩阵和向量当成如char,int,double一样的基本变量类型,为矩阵和向量提供了几乎是随心所欲的操作函数。因此,可以在此数值类库的基础上进行二次开发。类库的各项功能均经过严格的检测,并与MATLAB作了比较,结果准确无误,效率不相上下。书中以菜单的方式对数值分析类库的每项功能作了详尽的解释,给出了调用方法的示例,对某些算法还提供了相应的数学背景知识。 如果你经常需要使用计算机求解科学与工程中的数值计算问题,特别是希望使用VC++或BC++做出独立于MATLAB的应用软件,本类库是不错的选择。-err
Date
: 2026-01-06
Size
: 244kb
User
:
wt
[
Mathimatics-Numerical algorithms
]
1
DL : 0
用一个函数完成对sin cos exp 函数的不同上线积分-float integral(float (*p)(float),float a,float b,int n)
Date
: 2026-01-06
Size
: 10kb
User
:
杨玉丽
[
Mathimatics-Numerical algorithms
]
Conv-Int.h
DL : 0
Definitions of convolution interleaver cla-Definitions of convolution interleaver class
Date
: 2026-01-06
Size
: 1kb
User
:
Jawahar
[
Mathimatics-Numerical algorithms
]
gaosi
DL : 0
高斯列主元消去法 #include<stdio.h> #include<conio.h> #include<math.h> #define N 100 float a[N][N+1] void main( ) { int i,j,k,n float t,s=0, clrscr( ) printf("输入矩阵阶数:") scanf(" d",&n) printf("\n") printf("输入增广矩阵:\n") for(i=0 i<n i++) for(j=0 j<n+1 j++) scanf(" f",&a[i][j]) for(k=0 k<n-1 k++) ...... -高斯列主元消去法
Date
: 2026-01-06
Size
: 166kb
User
:
www.pudn.com
[
Mathimatics-Numerical algorithms
]
asd
DL : 0
用函数重载和带默认参数函数的方法写三个重载函数,求不同类型的两个数 的平方和。 要求: 1)用函数重载写三个重载函数,能分别求出两个整数,两个单精度和两个双精度数的平方和 2)其中int型重载函数中第一个参数的默认值为1,第二个参数的默认值为2 -By function overloading and default parameters function method with three overloaded functions to write, find different types of squares of two numbers. Requirements: 1) using function overloading to write three overloaded functions can be obtained, respectively, two integer, two single and two double-precision accuracy the number of square and 2) which overloaded function int type default first parameter value is 1, the second default value is 2
Date
: 2026-01-06
Size
: 3.96mb
User
:
刘梦露
[
Mathimatics-Numerical algorithms
]
zuixiao
DL : 0
最小重量机器设计问题,回溯法和分支限定法-#include<iostream> using namespace std #define N 3 #define M 3 int w[N][M]={{1,2,3},{3,2,1},{2,2,2}} int c[N][M]={{1,2,3},{3,2,1},{2,2,2}} class machine {public:
Date
: 2026-01-06
Size
: 7kb
User
:
林莹莹
[
Mathimatics-Numerical algorithms
]
hdu1297
DL : 0
杭电acm1297 #include<stdio.h> #include<string.h> #define m 1002 int f[m][70]={{1,1},{1,1},{1,2},{1,4}} void add(int p[],int q[],int sum[]) { int i,len if(p[0]<q[0])len=q[0] else len=p[0] for(i=1 i<=len i++) { sum[i]+=q[i]+p[i] if(sum[i]>=10000){sum[i]-=10000 sum[i+1]++ } }-acm contest
Date
: 2026-01-06
Size
: 222kb
User
:
韩飞
[
Mathimatics-Numerical algorithms
]
MAX-AND-MIN
DL : 0
输出各种数据类型的最大值、最小值。包括整形,长整形,浮点型等等。-output max and min,for example int,long.float,double and so on.
Date
: 2026-01-06
Size
: 265kb
User
:
戴呆
[
Mathimatics-Numerical algorithms
]
DirectIntBufferTest
DL : 0
Direct Int Buffer Test Source Code for Andriod.
Date
: 2026-01-06
Size
: 1kb
User
:
tcrketain
[
Mathimatics-Numerical algorithms
]
xzpjoz0
DL : 0
这是一个计算大数阶乘的算法. 主要是通过链表来保存结果的每一位数, 这样就不会有像long或int类型时,数太大而导致溢出()
Date
: 2026-01-06
Size
: 7kb
User
:
roinah
[
Mathimatics-Numerical algorithms
]
linkforsat
DL : 0
设计有理数类rational_number  数据成员: int numerator(分子) int den()
Date
: 2026-01-06
Size
: 1kb
User
:
Hagljg
[
Mathimatics-Numerical algorithms
]
795888
DL : 0
设计有理数类rational_number  数据成员: int numerator(分子) int den()
Date
: 2026-01-06
Size
: 1kb
User
:
leruggvd
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