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C/C++獲取輸入時的回車問題

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <malloc.h>

using namespace std;
void leftshift(char*a, int n, int m);

int main()
{
    char q = 0;
    do{
        char *a = (char *)malloc(20*sizeof(char));
        memset(a,0,20);
        int num = 0;
        int times = 0;
        if(q != 0)
            getchar();
//        cout << "Please input the string!" << endl;
        printf("Please input the string\n");
        gets(a);
        while(a[num] != 0)
        {
            printf("%c %d\n",a[num],num);
            num++;

        }
    //    while(scanf("%c",a+num) && a[num] != "\0")
    //    {
    //        num++;
    //    }
        cout << "The length of the input string " << num << endl;
        cout << "Please input the number you want to shift!" << endl;
        while(scanf("%d",×) && times <= 0)
        {
            //接受完輸入後,下次再獲取時,需要先將回車鍵獲取,以免下次獲取的內容中含回車鍵
            printf("Please input a positive number!\n");
            //標準輸出中的回車鍵跟這個不相關
        }
        leftshift(a,num,times);
        printf("After shift: %s\n",a);
        free(a);
        printf("Continue?y:n\n");
        getchar();  //之前有scanf獲取輸入,故此處先將回車獲取
//        do{
//            scanf("%c",&q);
//        }while(q!='y' || q!='n');
        scanf("%c",&q);

    }while(q=='y');
    return 0;
}


void leftshift(char*a,int n,int m)
{
    while(m--)
    {
        char tmp = a[0];
        for(int i = 1; i <= n; i++)
        {
            a[i-1] = a[i];
        }
        a[n-1] = tmp;
    }
}