亚洲欧美第一页_禁久久精品乱码_粉嫩av一区二区三区免费野_久草精品视频

蟲蟲首頁| 資源下載| 資源專輯| 精品軟件
登錄| 注冊

elEMents

  • While there are many textbooks about the European Union (EU), Clive Archer covers the essential el

    While there are many textbooks about the European Union (EU), Clive Archer covers the essential elEMents of the EU clearly and con- cisely, outlining the key debates and issues it faces today

    標(biāo)簽: essential textbooks the European

    上傳時(shí)間: 2013-12-18

    上傳用戶:498732662

  • While there are many textbooks about the European Union (EU), Clive Archer covers the essential el

    While there are many textbooks about the European Union (EU), Clive Archer covers the essential elEMents of the EU clearly and con- cisely, outlining the key debates and issues it faces today

    標(biāo)簽: essential textbooks the European

    上傳時(shí)間: 2014-01-07

    上傳用戶:qilin

  • While there are many textbooks about the European Union (EU), Clive Archer covers the essential el

    While there are many textbooks about the European Union (EU), Clive Archer covers the essential elEMents of the EU clearly and con- cisely, outlining the key debates and issues it faces today

    標(biāo)簽: essential textbooks the European

    上傳時(shí)間: 2013-12-17

    上傳用戶:zl5712176

  • While there are many textbooks about the European Union (EU), Clive Archer covers the essential el

    While there are many textbooks about the European Union (EU), Clive Archer covers the essential elEMents of the EU clearly and con- cisely, outlining the key debates and issues it faces today

    標(biāo)簽: essential textbooks the European

    上傳時(shí)間: 2013-12-09

    上傳用戶:stella2015

  • While there are many textbooks about the European Union (EU), Clive Archer covers the essential el

    While there are many textbooks about the European Union (EU), Clive Archer covers the essential elEMents of the EU clearly and con- cisely, outlining the key debates and issues it faces today

    標(biāo)簽: essential textbooks the European

    上傳時(shí)間: 2014-11-21

    上傳用戶:cursor

  • 24小時(shí)學(xué)通Qt編程

    Qt學(xué)習(xí)資料,界面設(shè)計(jì)必備。一個(gè)跨平臺(tái)的C++應(yīng)用程序開發(fā)框架。廣泛用于開發(fā)GUI程序,這種情況下又被稱為部件工具箱。也可用于開發(fā)非GUI程序,比如控制臺(tái)工具和服務(wù)器。Qt使用于OPIE、Skype、VLC media player、Adobe Photoshop elEMents、VirtualBox與Mathematica以及被Autodesk 、歐洲空間局、夢工廠、Google、HP、KDE、盧卡斯影業(yè)、西門子公司、富豪集團(tuán), 華特迪士尼動(dòng)畫制作公司、三星集團(tuán)、飛利浦、Panasonic 所使用。

    標(biāo)簽: Qt

    上傳時(shí)間: 2015-12-06

    上傳用戶:filling87

  • 《Qt及Qt Quick開發(fā)實(shí)戰(zhàn)精解》代碼

    《Qt及Qt Quick開發(fā)實(shí)戰(zhàn)精解》代碼。Qt(官方發(fā)音同cute 發(fā)音為/kju:t/,雖然也俗稱為Q.T.發(fā)音為/kju:ti?/")是一個(gè)跨平臺(tái)的C++應(yīng)用程序開發(fā)框架。廣泛用于開發(fā)GUI程序,這種情況下又被稱為部件工具箱。也可用于開發(fā)非GUI程序,比如控制臺(tái)工具和服務(wù)器。Qt使用于OPIE、Skype、VLC media player、Adobe Photoshop elEMents、VirtualBox與Mathematica以及被Autodesk 、歐洲空間局、夢工廠、Google、HP、KDE、盧卡斯影業(yè)、西門子公司、富豪集團(tuán), 華特迪士尼動(dòng)畫制作公司、三星集團(tuán)、飛利浦、Panasonic 所使用。

    標(biāo)簽: 《Qt及Qt Quick開發(fā)實(shí)戰(zhàn)精解》代碼

    上傳時(shí)間: 2015-12-06

    上傳用戶:filling87

  • 計(jì)算本征值程序

    Computes all eigenvalues and eigenvectors of a real symmetric matrix a, ! which is of size n by n, stored in a physical np by np array. ! On output, elEMents of a above the diagonal are destroyed. ! d returns the eigenvalues of a in its first n elEMents. ! v is a matrix with the same logical and physical dimensions as a, ! whose columns contain, on output, the normalized eigenvectors of a. ! nrot returns the number of Jacobi rotations that were required. ! Please notice that the eigenvalues are not ordered on output. ! If the sorting is desired, the addintioal routine "eigsrt" ! can be invoked to reorder the output of jacobi.

    標(biāo)簽: 計(jì)算 程序

    上傳時(shí)間: 2016-06-04

    上傳用戶:1512313

  • 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(biāo)簽: 算法 排序

    上傳時(shí)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(biāo)簽: c語言 算法 排序

    上傳時(shí)間: 2017-04-01

    上傳用戶:糖兒水嘻嘻

主站蜘蛛池模板: 湘潭市| 聂荣县| 光山县| 宜昌市| 文安县| 德清县| 稷山县| 当雄县| 彰化市| 鸡泽县| 那坡县| 宜兴市| 晋州市| 玛沁县| 西昌市| 叶城县| 尉犁县| 彭泽县| 盐边县| 微博| 孟村| 洮南市| 五莲县| 阿坝县| 青龙| 舞阳县| 凌源市| 托里县| 桃园县| 苏州市| 芮城县| 赤壁市| 集贤县| 三门峡市| 闽侯县| 北流市| 平和县| 当雄县| 壤塘县| 明光市| 普兰店市|