Divide and Conquer

     分治法，顾名思义：切分然后治理，治理就是各个解决问题然后合并整理。递归的解决分成n个较小规模的子问题，然后将各个结果合并从而解决原来的问题。
DIVIDE：将问题分解成一系列子问题。
CONQUER：递归地解决各个子问题，若问题足够小，则直接解决。
COMBINE：将子问题的结果合并成原问题的解。
自己编的代码，结果又运行不起来：ps:几经修改，看来细节问题要注意，简化推理是个好办法
//mergesort
#include<iostream>
using namespace std;

int merge(int*A,int p,int q,int r){
 int n1=q-p+1;
 int n2=r-q;
 int* L = new int[n1];
 int* R = new int[n2];
 for(int i=0;i<n1;++i)
  L[i]=A[p+i];
 for(int j=0;j<n2;++j)
  R[j]=A[q+1+j];
 i=0;j=0;
 for(int k=p;k<r;k++){
  if(L[i]<=R[j]){A[k]=L[i++];}
  else {A[k]=R[j++];}
 }
 return(0);
}

int mergesort(int*A,int p,int r){
 int q;
 if(p<r){
  q=(p+r)/2;
  mergesort(A,p,q);
  mergesort(A,q+1,r);
  merge(A,p,q,r);
 }
 return(0);
}

int main(){
    int b[6]={11,65,53,78,38,63};
    mergesort(b,0,5);
 for(int i=0;i<6;++i)
  cout<<b[i]<<'\t';
 return(0);
}

以下引用自：gooler的专栏http://blog.csdn.net/Gooler/archive/2006/03/22/632422.aspx有各种排序算法quicksort,heapsort...
/*
 * A is an array and p, q, and r are indices numbering elements of the array
 * such that p <= q < r.
 *
 * This procedure assumes that the subarrays A[p. .q] and A[q + 1. .r] are in
 * sorted order. It merges them to form a single sorted subarray that replaces
 * the current subarray A[p. .r].
 */
void merge(int A[], int p, int q, int r) {
    int i, j, k, size;
    int* B;
   
    size = r - p + 1;

    B = new int[size]; /* temp arry for storing the merge result */
   
    /* initialize B */
    for (i = 0; i < size; ++i) {
        B[i] = 0;
    }

    i = p;
    j = q + 1;
    k = 0;
   
    /* compare and copy the smaller to B */
    while (i <= q && j <= r) {
        if (A[i] < A[j]) {
            B[k++] = A[i++];
        } else {
            B[k++] = A[j++];
        }
    }
   
    /* copy the rest to B */
    while (i <= q) {
        B[k++] = A[i++];
    }   
    while (j <= r) {
        B[k++] = A[j++];
    }
   
    /* replace A[p..r] with B[0..r-p] */
    for (i = p, k = 0; i <= r; ++i, ++k) {
        A[i] = B[k];
    }

    delete[] B;
}

/*
 * This procedure sorts the elements in the subarray A[p. .r].
 *
 * If p >= r, the subarray has at most one element and is therefore
 * already sorted. Otherwise, the divide step simply computes an index
 * q that partitions A[p. .r] into two subarrays: A[p. .q], containing n/2
 * elements, and A[q + 1. .r], containing n/2 elements.
 */
void mergeSort(int A[], int p, int r) {
    if (p >= r) return;
   
    int q = (p + r) / 2;
   
    mergeSort(A, p, q);
    mergeSort(A, q + 1, r);
    merge(A, p, q, r);
}