题目描述：

#include 
     
      #include 
      
       using namespace std;/*  desc:complete knapsack problem, each items can select zero or more times.  auther: justinzhang(aktiger87@gmail.com)  time:2015年2月15日10:14:02  testcase#7:  input:  41 200012 105 92 20002 12 20001999 1999=======output:20001020001999=====input:31 656 83 3 3 3 3 39 109 4 4 9 4 9 9 9 9output:569*/#define MAXLEN  5000int main() {  int T = 0;   // The number of test cases;    int n = 0, k = 0;  // n is the number of integers, k is the volume of packages.    int **array =new int* [MAXLEN];  for(int i = 0; i< MAXLEN; i++) {    array[i] = new int[MAXLEN];  }    int tmp = 0;  cin>>T;    for(int i = 0; i < T; i++) {    vector
       
         arrayN;  // arrayN should be in this for loop, if it is a global one, you should clear it each time.    cin >> n; // the number of integers.    cin >> k; // the volume of package.    for(int j=0; j
        
         > tmp;      arrayN.push_back(tmp);    }        // here we need to construct the logic of select zero or more times for each item.    for(int j = 0; j < n; j++) {      int maxSelectNum = k/arrayN[j];      for(int select = 0; select < maxSelectNum; select++) {    int item = arrayN[j];    arrayN.push_back(item);          }          } // end of int j=0  , here we finish transition of complete knapsack problem to zeroone knapsack problem        //    cout << "size of arrayN is :" << arrayN.size() << endl;    for(int i=0; i
         
          = 0 ) {  // here we use >=0 , not >0, >0 will be an error, =0 filled the package.            v2 = array[vecIndex-1][volIndex-arrayN[vecIndex]] + arrayN[vecIndex];    }    if(v1 > v2 ) {      array[vecIndex][volIndex] = v1;    } else {      array[vecIndex][volIndex] = v2;    }      } // end of for loop volIndex     } // end of for loop int vecIndex        cout << array[arrayN.size()][k] << endl;  } // end of loop for int T  //释放动态申请的内存  for(int i =0; i < MAXLEN; i++) {    delete [] array[i];  }  delete[]array;  return 0;}