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Sum-Product

  • java學生數據庫

    /*import java.util.Scanner; //主類 public class student122 {   //主方法   public static void main(String[] args){     //定義7個元素的字符數組     String[] st = new String[7];     inputSt(st);       //調用輸入方法     calculateSt(st);   //調用計算方法     outputSt(st);      //調用輸出方法   }   //其他方法   //輸入方法 private static void inputSt(String st[]){     System.out.println("輸入學生的信息:");   System.out.println("學號 姓名 成績1,2,3");   //創建鍵盤輸入類   Scanner ss = new Scanner(System.in);   for(int i=0; i<5; i++){     st[i] = ss.next(); //鍵盤輸入1個字符串   } }   //計算方法 private static void calculateSt(String[] st){   int sum = 0;         //總分賦初值 int ave = 0;         //平均分賦初值 for(int i=2;i<5;i++) {   /計總分,字符變換成整數后進行計算   sum += Integer.parseInt(st[i]); } ave = sum/3;         //計算平均分 //整數變換成字符后保存到數組里 st[5] = String.valueOf(sum); st[6] = String.valueOf(ave); }   //輸出方法 private static void outputSt(String[] st){     System.out.print("學號 姓名 ");   //不換行   System.out.print("成績1 成績2 成績3 ");   System.out.println("總分 平均分");//換行   //輸出學生信息   for(int i=0; i<7; i++){     //按格式輸出,小于6個字符,補充空格     System.out.printf("%6s", st[i]);   }   System.out.println();            //輸出換行 } }*/   import java.util.Scanner;   public class student122 {   public static void main(String[] args) { // TODO 自動生成的方法存根 String[][] st = new String[3][8]; inputSt(st); calculateSt(st); outputSt(st); }   //輸入方法 private static void inputSt(String st[][]) { System.out.println("輸入學生信息:"); System.out.println("班級 學號 姓名 成績:數學 物理 化學"); //創建鍵盤輸入類 Scanner ss = new Scanner(System.in); for(int j = 0; j < 3; j++) { for(int i = 0; i < 6; i++) { st[j][i] = ss.next(); } } } //輸出方法 private static void outputSt(String st[][]) { System.out.println("序號 班級 學號 姓名 成績:數學 物理 化學 總分 平均分"); //輸出學生信息 for(int j = 0; j < 3; j++) { System.out.print(j+1 + ":"); for(int i = 0; i < 8; i++) { System.out.printf("%6s", st[j][i]); } System.out.println(); } }     //計算方法     private static void calculateSt(String[][] st)     {      int sum1 = 0;      int sum2 = 0; int sum3 = 0;      int ave1 = 0;      int ave2 = 0;      int ave3 = 0;      for(int i = 3; i < 6; i++)      {      sum1 += Integer.parseInt(st[0][i]);      }      ave1 = sum1/3;           for(int i = 3; i < 6; i++)      {      sum2 += Integer.parseInt(st[1][i]);      }      ave2 = sum2/3;           for(int i = 3; i < 6; i++)      {      sum3 += Integer.parseInt(st[2][i]);      }      ave3 = sum3/3;           st[0][6] = String.valueOf(sum1);      st[1][6] = String.valueOf(sum2);      st[2][6] = String.valueOf(sum3);      st[0][7] = String.valueOf(ave1);      st[1][7] = String.valueOf(ave2);      st[2][7] = String.valueOf(ave3);     } }

    標簽: java 數據庫

    上傳時間: 2017-03-17

    上傳用戶:simple

  • C語言算法排序問題

    1.Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose sum is exactly x. (Implement exercise 2.3-7.)

    標簽: 算法 排序

    上傳時間: 2017-04-01

    上傳用戶:糖兒水嘻嘻

  • c語言算法排序

    1.Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose sum is exactly x. (Implement exercise 2.3-7.) #include<stdio.h> #include<stdlib.h> void merge(int arr[],int low,int mid,int high){      int i,k;      int *tmp=(int*)malloc((high-low+1)*sizeof(int));      int left_low=low;      int left_high=mid;      int right_low=mid+1;      int right_high=high;      for(k=0;left_low<=left_high&&right_low<=right_high;k++)      {      if(arr[left_low]<=arr[right_low]){                                        tmp[k]=arr[left_low++];                                        }      else{           tmp[k]=arr[right_low++];           } }             if(left_low<=left_high){                              for(i=left_low;i<=left_high;i++){                                                               tmp[k++]=arr[i];                                                               }                              }       if(right_low<=right_high){                              for(i=right_low;i<=right_high;i++)                                                                tmp[k++]=arr[i];                                                        }                              for(i=0;i<high-low+1;i++)                                                       arr[low+i]=tmp[i];       } void merge_sort(int a[],int p,int r){      int q;      if(p<r){              q=(p+r)/2;              merge_sort(a,p,q);              merge_sort(a,q+1,r);              merge(a,p,q,r);              }      } int main(){     int a[8]={3,5,8,6,4,1,1};     int i,j;     int x=10;     merge_sort(a,0,6);     printf("after Merging-Sort:\n");     for(i=0;i<7;i++){                      printf("%d",a[i]);                      }     printf("\n");     i=0;j=6;     do{                                    if(a[i]+a[j]==x){                                  printf("exist");                                  break;                                  }                  if(a[i]+a[j]>x)                                 j--;                  if(a[i]+a[j]<x)                                 i++;                       }while(i<=j);     if(i>j)              printf("not exist");     system("pause");     return 0;     }

    標簽: c語言 算法 排序

    上傳時間: 2017-04-01

    上傳用戶:糖兒水嘻嘻

  • 有限差分法

    function [alpha,N,U]=youxianchafen2(r1,r2,up,under,num,deta)      %[alpha,N,U]=youxianchafen2(a,r1,r2,up,under,num,deta)   %該函數用有限差分法求解有兩種介質的正方形區域的二維拉普拉斯方程的數值解   %函數返回迭代因子、迭代次數以及迭代完成后所求區域內網格節點處的值   %a為正方形求解區域的邊長   %r1,r2分別表示兩種介質的電導率   %up,under分別為上下邊界值   %num表示將區域每邊的網格剖分個數   %deta為迭代過程中所允許的相對誤差限      n=num+1; %每邊節點數   U(n,n)=0; %節點處數值矩陣   N=0; %迭代次數初值   alpha=2/(1+sin(pi/num));%超松弛迭代因子   k=r1/r2; %兩介質電導率之比   U(1,1:n)=up; %求解區域上邊界第一類邊界條件   U(n,1:n)=under; %求解區域下邊界第一類邊界條件   U(2:num,1)=0;U(2:num,n)=0;      for i=2:num   U(i,2:num)=up-(up-under)/num*(i-1);%采用線性賦值對上下邊界之間的節點賦迭代初值   end   G=1;   while G>0 %迭代條件:不滿足相對誤差限要求的節點數目G不為零   Un=U; %完成第n次迭代后所有節點處的值   G=0; %每完成一次迭代將不滿足相對誤差限要求的節點數目歸零   for j=1:n   for i=2:num   U1=U(i,j); %第n次迭代時網格節點處的值      if j==1 %第n+1次迭代左邊界第二類邊界條件   U(i,j)=1/4*(2*U(i,j+1)+U(i-1,j)+U(i+1,j));   end         if (j>1)&&(j                 U2=1/4*(U(i,j+1)+ U(i-1,j)+ U(i,j-1)+ U(i+1,j));    U(i,j)=U1+alpha*(U2-U1); %引入超松弛迭代因子后的網格節點處的值      end      if i==n+1-j %第n+1次迭代兩介質分界面(與網格對角線重合)第二類邊界條件   U(i,j)=1/4*(2/(1+k)*(U(i,j+1)+U(i+1,j))+2*k/(1+k)*(U(i-1,j)+U(i,j-1)));      end      if j==n %第n+1次迭代右邊界第二類邊界條件   U(i,n)=1/4*(2*U(i,j-1)+U(i-1,j)+U(i+1,j));   end   end   end   N=N+1 %顯示迭代次數   Un1=U; %完成第n+1次迭代后所有節點處的值   err=abs((Un1-Un)./Un1);%第n+1次迭代與第n次迭代所有節點值的相對誤差   err(1,1:n)=0; %上邊界節點相對誤差置零   err(n,1:n)=0; %下邊界節點相對誤差置零    G=sum(sum(err>deta))%顯示每次迭代后不滿足相對誤差限要求的節點數目G   end

    標簽: 有限差分

    上傳時間: 2018-07-13

    上傳用戶:Kemin

  • C語言編寫雅可比迭代

    # include<stdio.h> # include<math.h> # define N 3 main(){     float NF2(float *x,float *y);     float A[N][N]={{10,-1,-2},{-1,10,-2},{-1,-1,5}};     float b[N]={7.2,8.3,4.2},sum=0;     float x[N]= {0,0,0},y[N]={0},x0[N]={};     int i,j,n=0;     for(i=0;i<N;i++)     {         x[i]=x0[i];     }     for(n=0;;n++){                 //計算下一個值     for(i=0;i<N;i++){         sum=0;         for(j=0;j<N;j++){             if(j!=i){                 sum=sum+A[i][j]*x[j];             }         }         y[i]=(1/A[i][i])*(b[i]-sum);         //sum=0;     }     //判斷誤差大小         if(NF2(x,y)>0.01){                 for(i=0;i<N;i++){         x[i]=y[i];     }     }             else             break;     }     printf("經過%d次雅可比迭代解出方程組的解:\n",n+1);     for(i=0;i<N;i++){         printf("%f      ",y[i]);     } } //求兩個向量差的二范數函數 float NF2(float *x,float *y){ int i; float z,sum1=0; for(i=0;i<N;i++){     sum1=sum1+pow(y[i]-x[i],2); } z=sqrt(sum1); return z; }

    標簽: C語言 編寫 迭代

    上傳時間: 2019-10-13

    上傳用戶:大萌萌撒

  • Delphi三層數據庫連接池 (1)

    Delphi三層數據庫連接池 (1) Connection Pool for Delphi release notes ------------------------------------------------------------------------------- This document contains: - Short description of the product - Other text files - TRIAL version limitations - Delphi - versions supported - Installation of Connection Pool for Delphi - Installation of Connection Pool for Delphi help file - Ordering information - Support and Web resources - Thanks To

    標簽: Delphi 數據庫 連接

    上傳時間: 2019-12-12

    上傳用戶:me2008

  • AO4420

    AO4420, AO4420L ( Green Product ) N-Channel Enhancement Mode Field Effect Transistor

    標簽: 4420 AO

    上傳時間: 2020-04-19

    上傳用戶:su1254

  • AD810

    PRODUCT DESCRIPTION The AD810 is a composite and HDTV compatible, current feedback, video operational amplifier, ideal for use in systems such as multimedia, digital tape recorders and video cameras. The 0.1 dB flatness specification at bandwidth of 30 MHz (G = +2) and the differential gain and phase of 0.02% and 0.04° (NTSC) make the AD810 ideal for any broadcast quality video system. All these specifications are under load conditions of 150 ? (one 75 ? back terminated cable). The AD810 is ideal for power sensitive applications such as video cameras, offering a low power supply current of 8.0 mA max. The disable feature reduces the power supply current to only 2.1 mA, while the amplifier is not in use, to conserve power. Furthermore the AD810 is specified over a power supply range of ±5 V to ±15 V.

    標簽: 810 AD

    上傳時間: 2020-04-19

    上傳用戶:su1254

  • 4G & Beyond Convergence of Networks

    From the transition of analog to digital communication along with seamless mobility and high computing power of small handheld devices, the wireless communications industry has seen tremendous changes leading to the integration of several telecommunication networks, devices and services over last 30 years. The rate of this progress and growth has increased particularly in the past decade because people no longer use their devices and networks for voice only, but demand bundle contents such as data download/streaming, HDTV, HD video , 3D video conferencing with higher efficiency, seamless connectivity, intelligence, reliability and better user experience. Although the challenges facing service providers and telecommunication companies differ by product, region, market size, and their areas of concentration but time to market, efficient utilization of their assets and revenue expansion, have impacted significantly how to manage and conduct their business while maintaining sufficient margin. 

    標簽: Convergence Networks Beyond 4G of

    上傳時間: 2020-05-26

    上傳用戶:shancjb

  • SDxCentral-Mega-NFV-Report

    The following sections profile many of the vendors in the NFV market. The individual profiles were created through a collaborative effort between SDNCentral’s Research Team and the Vendor’s product experts. SDNCentral worked under the assumption the information provided by the vendors was factual, auditing the submissions only to remove unverifiable claims and hyperbole. Extended profiles can be viewed online. While every attempt has been made to validate the capabilities listed in the profiles, SDNCentral advises end users to verify the veracity of each claim for themselves in their actual deployment environments. SDNCentral cannot be held liable for unexpected operations, damages or incorrect operation due to any inaccuracies listed here. SDNCentral welcomes feedback and additional information from end users based on their real-world experiences with the products and technologies listed. The SDNCentral research team can be reached at

    標簽: SDxCentral-Mega-NFV-Report

    上傳時間: 2020-06-01

    上傳用戶:shancjb

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