C++算法设计中的组合问题

分治法解决最大子段和问题

int MaxSum(int a[ ], int left, int right)
   {
       int sum=0,midsum=0,leftsum=0,rightsum=0;
       int center,s1,s2,lefts,rights;
       if (left==right) {      //如果序列长度为1，直接求解
           sum=a[left];          
       }
      else {
          center=(left+right)/2;    //划分
          leftsum=MaxSum(a, left, center);  
                                                       //对应情况①，递归求解
          rightsum=MaxSum(a, center+1, right);         //对应情况②，递归求解                                                     
        s1=0; lefts=0;             //以下对应情况③，先求解s1
        for (int i=center; i>=left; i--)
        {
            lefts+=a[i];
            if (lefts>s1) s1=lefts;
        }
        s2=0; rights=0;             //再求解s2
        for (int j=center+1; j<=right; j++)
        { 
            rights+=a[j];
            if (rights>s2) s2=rights;
        }
        sum=s1+s2;              //计算情况③的最大子段和 
        if (sum

减治法解决淘汰赛冠军问题

bool Comp(char A,char B){//A大就赢 
   if(A1) {
	    i=i/2;
        for (j=0; j

减治法和递归解决假币问题
假设：假币比较轻

int a[]={2,2,2,2,2,1,2,2}; 
int Coin(int low,int high,int n){
	int i,num1,num2,num3;
	int add1=0,add2=0;
	if(n==1)  return low+1;//返回序号，故下标加一
	if(n%3==0)  num1=num2=n/3;
	else  num1=num2=n/3+1;
	num3=n-num1-num2;
	for(i=0;iadd2)   return Coin(low+num1,low+num1+num2-1,num2);
	else  return Coin(low+num1+num2,high,num3);
}

测试：

int main(){
	int a[]={-20,11,-4,13,-5,-2};
	cout<