并行高性能计算——找最大值和最小值

#include<iostream>
using namespace std;
struct nums{
int values[2];
};
static nums realVal;
static nums formVal;
struct nums getVal(int val1,int val2){
	struct nums myvalue;
	if(val1>val2){
		myvalue.values[0]=val1;
		myvalue.values[1]=val2;
	}
	else{
		myvalue.values[0]=val2;
		myvalue.values[1]=val1;
	}
	return myvalue;
}
struct nums recurfind(int A[],int low,int high){	
	if(low==high){
		formVal.values[0]=A[low];
		formVal.values[1]=A[low];
		return formVal;
	}
	else if(low==high-1){
		formVal=getVal(A[low],A[high]);
		return formVal;
	}
	struct nums tempVal1=recurfind(A,low,(low+high)/2);
	struct nums tempVal2=recurfind(A,(low+high)/2+1,high);
	if(tempVal1.values[0]>tempVal2.values[0])
		formVal.values[0]=tempVal1.values[0];
	else
		formVal.values[0]=tempVal2.values[0];
	if(tempVal1.values[1]<tempVal2.values[1])
		formVal.values[1]=tempVal1.values[1];
	else
		formVal.values[1]=tempVal2.values[1];
	return formVal;
}

int main(){
	int A[]={2,5,13,9,4,12,8};
	int low=0;
	int high=6;
	realVal=recurfind(A,low,high);
	cout<<realVal.values[0]<<" "<<realVal.values[1]<<endl;
	return 0;
}

时间复杂度：T(n)=2T(n/2)+2=4T(n/4)+2²+2=...=2^KT(n/2^K)+2(1-2^K)/(1-2);有T(2)=1,此时n=2^K+1,则T(n)=(n/2)2+n-2=1.5n-2