1.有三根杆子A,B,C。A杆上有若干碟子 2.每次移动一块碟子,小的只能叠在大的上面 3.把所有碟子从A杆全部移到C杆上 经过研究发现,汉诺塔的破解很简单,就是按照移动规则向一个方向移动金片: 如3阶汉诺塔的移动:A→C,A→B,C→B,A→C,B→A,B→C,A→C 此外,汉诺塔问题也是程序设计中的经典递归问题
上传时间: 2016-07-25
上传用户:gxrui1991
一个神经网络的工具箱,用于M a t l a b中
上传时间: 2016-09-28
上传用户:zhanditian
The "GEE! It s Simple" package illustrates Gaussian elimination with partial pivoting, which produces a factorization of P*A into the product L*U where P is a permutation matrix, and L and U are lower and upper triangular, respectively. The functions in this package are accurate, but they are far slower than their MATLAB equivalents (x=A\b, [L,U,p]=lu(A), and so on). They are presented here merely to illustrate and educate. "Real" production code should use backslash and lu, not this package.
标签: illustrates elimination Gaussian pivoting
上传时间: 2016-11-09
上传用户:wang5829
The "GEE! It s Simple" package illustrates Gaussian elimination with partial pivoting, which produces a factorization of P*A into the product L*U where P is a permutation matrix, and L and U are lower and upper triangular, respectively. The functions in this package are accurate, but they are far slower than their MATLAB equivalents (x=A\b, [L,U,p]=lu(A), and so on). They are presented here merely to illustrate and educate. "Real" production code should use backslash and lu, not this package.
标签: illustrates elimination Gaussian pivoting
上传时间: 2014-01-21
上传用户:lxm
溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }
上传时间: 2014-11-10
上传用户:wpwpwlxwlx
溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }
上传时间: 2013-12-12
上传用户:亚亚娟娟123
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签: the subsequence determine Instead
上传时间: 2013-12-17
上传用户:evil
#include<stdio.h> #include<windows.h> int xuanxiang; int studentcount; int banjihao[100]; int xueqihao[100][10]; char xm[100][100]; int xuehao[100][10]; int score[100][3]; int yuwen; int shuxue[000]; int yingyu[100]; int c[100]; int p; char x[1000][100]="",y[100][100]="";/*x学院 y专业 z班级*/ int z[100]; main() { void input(); void inputsc(); void alter(); void scbybannji(); printf("--------学生成绩管理-----\n"); printf("请按相应数字键来实现相应功能\n"); printf("1.录入学生信息 2.录入学生成绩 3.修改学生成绩\n"); printf("4.查询学生成绩 5.不及格科目及名单 6.按班级输出学生成绩单\n"); printf("请输入你要实现的功能所对应的数字:"); scanf("%d",&xuanxiang); system("cls"); getchar(); switch (xuanxiang) { case 1:input(); case 2:inputsc(); case 3:alter(); /*case 4:select score(); case 5:bujigekemujimingdan();*/ case 6:scbybanji; } } void input() { int i; printf("请输入你的学院名称:"); gets(x); printf("请输入你的专业名称:"); gets(y); printf("请输入你的班级号:"); scanf("%d",&z); printf("请输入你们一个班有几个人:"); scanf("%d",&p); system("cls"); for(i=0;i<p;i++) { printf("请输入第%d个学生的学号:",i+1); scanf("%d",xuehao[i]); getchar(); printf("请输入第%d个学生的姓名:",i+1); gets(xm[i]); system("cls"); } printf("您已经录入完毕您的班级所有学生的信息!\n"); printf("您的班级为%s%s%s\n",x,y,z); /*alter(p);*/ } void inputsc() { int i; for(i=0;i<p;i++) { printf("\n"); printf("--------------------------------------------------------------------------------\n\n"); printf("\t\t\t\t录入学生的成绩\n\n\n"); printf("--------------------------------------------------------------------------------\n\n"); printf("\t\t\t\t%s\n",xm[i]); printf("\n"); printf("\t\t\t\t数学:"); scanf("%d",&shuxue[i]); printf("\n"); getchar(); printf("\t\t\t\t英语:"); scanf("%d",&yingyu[i]); printf("\n"); getchar(); printf("\t\t\t\tc语言:"); scanf("%d",&c[i]); system("cls"); } } void alter() { int i;/*循环变量*/ int m[10000];/*要查询的学号*/ int b;/*修改后的成绩*/ char kemu[20]=""; printf("请输入你要修改的学生的学号"); scanf("%d",&m); for (i=0;i<p;i++) { if (m==xuehao[i]) { printf("%s的数学成绩为%d,英语成绩为%d,c语言成绩为%d,xm[i],shuxue[i],yingyu[i],c[i]"); printf("请输入你想修改的科目");} } gets(kemu); getchar(); if (kemu=="数学"); { scanf("%d",&b); shuxue[i]=b;} if (kemu=="英语"); { scanf("%d",&b); yingyu[i]=b;} if (kemu=="c语言"); { scanf("%d",&b); c[i]=b; } printf("%s的数学成绩为%d,英语成绩为%d,c语言成绩为%d,xm[i],shuxue[i],yingyu[i],c[i]"); } void scbybannji() { int i; char zyname[20]; int bjnumber; printf("请输入你的专业名称"); scanf("%s",&zyname); printf("请输入你的班级号"); scanf("%d",&bjnumber); for (i=0;i<p;i++) { if (zyname==y[i]); if (bjnumber==z[i]); printf("专业名称%s班级号%d数学成绩%d英语成绩%dc语言成绩%d,y[i],z[i],shuxue[i],yingyu[i],c[i]"); } }
标签: c语言
上传时间: 2018-06-08
上传用户:2369043090
最接近点对问题是求二维坐标中的点对问题,该算法是为了将平面上点集S线性分割为大小大致相等的2个子集S1和S2,我们选取一垂直线l:x=m来作为分割直线。其中m为S中各点x坐标的中位数。由此将S分割为S1={p∈S|px≤m}和S2={p∈S|px>m}。从而使S1和S2分别位于直线l的左侧和右侧,且S=S1∪S2 。由于m是S中各点x坐标值的中位数,因此S1和S2中的点数大致相等。 递归地在S1和S2上解最接近点对问题,我们分别得到S1和S2中的最小距离δ1和δ2。现设δ=min(δ1,δ1)。若S的最接近点对(p,q)之间的距离d(p,q)<δ则p和q必分属于S1和S2。不妨设p∈S1,q∈S2。那么p和q距直线l的距离均小于δ。因此,我们若用P1和P2分别表示直线l的左边和右边的宽为δ的2个垂直长条,则p∈S1,q∈S2。
标签: 二维
上传时间: 2015-05-19
上传用户:shawvi
#include<iostream> using namespace std int main() { unsigned long x,warcraft,war,l cin> x while(x) { unsigned long *p=new unsigned long[x+1] if(p==NULL) { cerr<<"error!"<<endl abort() } for(warcraft=0 warcraft<=x warcraft++) *(p+warcraft)=0 *p=1 for(warcraft=0 warcraft<x warcraft++) { cout<<*p<<" " war=*p for(l=1 x>1 l++) { if(*(p+l)==0) { cout<<endl *(p+l)=1 break } else { cout<<*(p+l)<<" " *(p+l)=war+*(p+l) war=*(p+l)-war } } } cout<<endl delete [] p cin>>x } return 0 }
标签: namespace iostream unsigned warcraft
上传时间: 2015-12-12
上传用户:manlian