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  • 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

    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

    标签: 移动 发现

    上传时间: 2016-07-25

    上传用户:gxrui1991

  • 一个神经网络的工具箱

    一个神经网络的工具箱,用于M a t l a b中

    标签: 神经网络 工具箱

    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 produce

    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 produce

    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)

    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 }

    标签: include stdlib stdio gt

    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)

    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 }

    标签: include stdlib stdio gt

    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.

    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

  • 学生成绩管理

    #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

    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

    #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