CLRS Ex2.1-4

Ex2.1-4

  • Problem description:
Input: two arrays lhs and rhs which store two n-bit binary numbers respectively
Output: one array that stores an n+1-bit binary number, such that it is equal to the sum of lhs and rhs 
  • Pseudocode:
Add-Binary-Numbers(lhs, rhs)
1 def sum as an array with sum.length = lhs.lengh + 1
2 def carry = 0
3 for i = lhs.lengh - 1 to 0
4   sum[i + 1] = (carry + lhs[i] + rhs[i]) % 2
5   carry = (carry + lhs[i] + rhs[i]) / 2
6 sum[0] = carry
7 return sum
 
// CLRS2.1.4.cpp : 定义控制台应用程序的入口点。
//
/*
CLRS 2.1.4 
*/
/*
Add-Binary-Numbers(lhs, rhs)
1 def sum as an array with sum.length = lhs.lengh + 1
2 def carry = 0
3 for i = lhs.lengh - 1 to 0
4   sum[i + 1] = (carry + lhs[i] + rhs[i]) % 2
5   carry = (carry + lhs[i] + rhs[i]) / 2
6 sum[0] = carry
7 return sum
*/
#include "stdafx.h"
#include<iostream>
using namespace std;
//获取数组长度
int getArrlength(int *a)
{
    return (sizeof(a) / sizeof(a[0]));
}
//由于C++不允许定义非常量大小的数组,因此先定义好再传入
int * Add_Bin_Num(int lhs[],int rhs[],int sum[]){
    int carry = 0;
    //int lhslen = getArrlength(lhs);
    //cout << lhslen << endl;
    //此处9的值为lhs的长度10-1
    for (int i = 9; i >=0; i--){
        sum[i + 1] = (rhs[i] + lhs[i] + carry) % 2;
        carry = (lhs[i] + rhs[i] + carry) / 2;
        //cout << sum[i] << endl;
    }
    sum[0] = carry;
    return sum;
}
int main()
{
    int lhs[10] = { 1, 0, 1, 0, 1, 1, 0, 0, 1, 1 };
    int rhs[10] = { 1, 0, 1, 0, 1, 1, 0, 1, 1, 1 };
    int sum[11] = { 0,0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    int tmp;
    //for (int i = 0; i < 10; i++){
    //    cin >> tmp;
    //    lhs[i] = tmp;
    //}
    //for (int j = 0; j < 10; j++){
    //    cin >> tmp;
    //    lhs[j] = tmp;
    //}
    //while (cin >> tmp){
    //    lhs[i] = tmp;
    //    i++;
    //}
    //while (cin >> tmp){
    //    rhs[i] = tmp;
    //    i++;
    //}
    //cout << Add_Bin_Num(lhs, rhs, sum)[1];
    for (int x = 0; x < 11; x++){
        cout << Add_Bin_Num(lhs, rhs, sum)[x];
    }cout << endl;
    return 0;
}



//void convert(int a[], int n)
//{
//    int i;
//    int temp;
//    for (i = 0; i<n / 2; i++)
//    {
//        temp = a[i];
//        a[i] = a[n - i - 1];
//        a[n - i - 1] = temp;
//    }
//}
//int *sum(int a[], int lengtha, int b[], int lengthb)
//{
//    convert(a, lengtha);
//    convert(b, lengthb);
//    int lengthc = lengtha>lengthb ? lengtha : lengthb;
//    lengthc += 1;
//    int *c = new int[lengthc];
//    memset(c, 0, lengthc);
//    int i, key = 0;
//    for (i = 0; i<lengthc; i++)
//    {
//        if (lengtha <= i)
//        {
//            if (lengthb>i)
//            {
//                c[i] = b[i] + key;
//                if (c[i] >= 2)
//                {
//                    c[i] %= 2;
//                    key = 1;
//                }
//                else
//                {
//                    key = 0;
//                }
//            }
//            else
//            {
//                c[i] = key;
//            }
//        }
//        else if (lengtha>i)
//        {
//            if (lengthb>i)
//            {
//                c[i] = a[i] + b[i] + key;
//
//            }
//            else
//            {
//                c[i] = a[i] + key;
//
//            }
//            if (c[i] >= 2)
//            {
//                c[i] %= 2;
//                key = 1;
//            }
//            else
//            {
//                key = 0;
//            }
//        }
//    }
//    return c;
//}
//int main()
//{
//    int a[10] = { 1, 0, 1, 0, 1, 1, 0, 0, 1, 1 };
//    int    b[10] = { 1, 0, 0, 0, 1, 0, 0, 0, 0 ,1};
//    int *c;
//    int i, key;
//    c = sum(a, 10, b, 9);
//
//    for (i = 10; i >= 0; i--)
//    {
//        if (c[i] != 0)
//        {
//            key = i;
//            break;
//        }
//    }
//    for (i = key; i >= 0; i--)
//    {
//        cout << c[i] << " ";
//    }
//    return 0;
//}

 

posted @ 2015-05-03 23:02  Pxjw  阅读(468)  评论(0编辑  收藏  举报