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.
標簽： represented integers group items
上傳時間： 2016-01-17
上傳用戶：jeffery
The XML Toolbox converts MATLAB data types (such as double, char, struct, complex, sparse, logical) of any level of nesting to XML format and vice versa. For example, >> project.name = MyProject >> project.id = 1234 >> project.param.a = 3.1415 >> project.param.b = 42 becomes with str=xml_format(project, off ) "<project> <name>MyProject</name> <id>1234</id> <param> <a>3.1415</a> <b>42</b> </param> </project>" On the other hand, if an XML string XStr is given, this can be converted easily to a MATLAB data type or structure V with the command V=xml_parse(XStr).
標簽： converts Toolbox complex logical
上傳時間： 2016-02-12
上傳用戶：a673761058
寫一個對文本文件加密的程序和一個解密的程序。密碼規則是：對于小寫字母，a換成x，b換成y，c換成z，d換成a，e換成b，...；對于大寫字母，A換成X，B換成Y，C換成Z，D換成A，E換成B，...；其他字符不變。
上傳時間： 2016-08-16
上傳用戶：jennyzai
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標簽： AOrigin APoint Point PointToAngle
上傳時間： 2016-10-31
上傳用戶：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
標簽： AOrigin APoint Point PointToAngle
上傳時間： 2013-12-18
上傳用戶：rocketrevenge
漢諾塔！！！ Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation eg. if n = 2 A→B A→C B→C if n = 3 A→C A→B C→B A→C B→A B→C A→C
標簽： the animation Simulate movement
上傳時間： 2017-02-11
上傳用戶：waizhang
實驗源代碼 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("請輸入矩陣第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可傳遞閉包關系矩陣是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元關系的可傳遞閉包\n"); void warshall(int,int); int k , n; printf("請輸入矩陣的行數 i: "); scanf("%d",&k); 四川大學實驗報告 printf("請輸入矩陣的列數 j: "); scanf("%d",&n); warshall(k,n); }
上傳時間： 2016-06-27
上傳用戶：梁雪文以
題目：古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數為多少？ //這是一個菲波拉契數列問題 public class lianxi01 { public static void main(String[] args) { System.out.println("第1個月的兔子對數: 1"); System.out.println("第2個月的兔子對數: 1"); int f1 = 1, f2 = 1, f, M=24; for(int i=3; i<=M; i++) { f = f2; f2 = f1 + f2; f1 = f; System.out.println("第" + i +"個月的兔子對數: "+f2); } } } 【程序2】 題目：判斷101-200之間有多少個素數，并輸出所有素數。 程序分析：判斷素數的方法：用一個數分別去除2到sqrt(這個數)，如果能被整除， 則表明此數不是素數，反之是素數。 public class lianxi02 { public static void main(String[] args) { int count = 0; for(int i=101; i<200; i+=2) { boolean b = false; for(int j=2; j<=Math.sqrt(i); j++) { if(i % j == 0) { b = false; break; } else { b = true; } } if(b == true) {count ++;System.out.println(i );} } System.out.println( "素數個數是: " + count); } } 【程序3】 題目：打印出所有的 "水仙花數 "，所謂 "水仙花數 "是指一個三位數，其各位數字立方和等于該數本身。例如：153是一個 "水仙花數 "，因為153=1的三次方＋5的三次方＋3的三次方。 public class lianxi03 { public static void main(String[] args) { int b1, b2, b3;
上傳時間： 2017-12-24
上傳用戶：Ariza
#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計算U得:"<<endl; U.Disp(); cout<<"計算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
標簽： 道理特分解法
上傳時間： 2018-05-20
上傳用戶：Aa123456789
