代码搜索:有限差分

找到约 10,000 项符合「有限差分」的源代码

代码结果 10,000
www.eeworm.com/read/480097/1323054

txt 郑凌-6分.txt

//圆排列的最小长度算法 040000003 郑凌 #include #include #include #include //==================随机数类=========================== const unsigned long maxshort=65
www.eeworm.com/read/480097/1323056

txt 林政-3.5分.txt

#include #include #include #include class circle { private: float min; //当前最优值 float *x; //当前圆排列圆心横坐标 float *r; //当前圆排列 int n;
www.eeworm.com/read/480097/1323057

txt 梁淘-3分.txt

#include #include #include class triangle { friend int compute(int);//私有变量初始化函数 private: void backtrack(int); int n, //第一行的符号个数 half, //n*(n+
www.eeworm.com/read/480097/1323060

txt 孙锋-6分.txt

//圆排列的随机化算法 #include #include #include #include //随机数类 const unsigned long maxshort=65536L; const unsigned long multiplier=1194211693L; const u
www.eeworm.com/read/480097/1323061

txt 王敏-2.5分.txt

#include #include #include #include #include #include #include long n;//棋盘的大小 int num;//可放置战车数 char * * chessBoar
www.eeworm.com/read/480097/1323063

txt 孙锋-6分.txt

#include #include #include int n; int *data; void readData() { ifstream inStream; inStream.open("input.txt"); if(!inStream) { cerr
www.eeworm.com/read/480097/1323067

txt 郑凌-4.5分.txt

//计算程序的最优存储方案 #include #include using namespace std; float MinTime(int, float*, float*); float MinTime(int n, float *l, float *p) { int i, j; float a=0, b=0, *t;
www.eeworm.com/read/480097/1323068

txt 周恒-6分.txt

#include #include #include char *x,*y,**b; int **c; ofstream fout("output.txt"); void LCSLength(int m,int n,char *x,char *y,int **c,char **b) { int i,j
www.eeworm.com/read/480097/1323070

txt 陈星娥-6分.txt

#include < iostream.h > #include < fstream.h > #include < stdio.h > ofstream myoutf("output.txt"); //输出数据到文件, void LISlength(int m,int * x,int * y); void main() { int size; //定义序列的长度 i
www.eeworm.com/read/480097/1323075

txt 谢富平-6分.txt

#include < stdio.h > #include < fstream.h > #include < iostream.h > ofstream myoutf("output.txt"); int Max_Sum(int m,int n, int * a) { if (n < m || m < 1) return 0; int * b = new int[n +