HDOJ 2051-2060

标签：2051 case int number base each 2060 Input HDOJ

2050Bitset

Problem Description Give you a number on base ten,you should output it on base two.(0 < n < 1000)   Input For each case there is a postive number n on base ten, end of file.   Output For each case output a number on base two.   Sample Input 1 2 3   Sample Output 1 10 11 人话：十进制转二进制

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    int a[16];
    int i, j;
    while (scanf("%d", &n) != EOF)
    {
        i = 0;
        while (n > 0)
        {
            a[i] = n % 2;
            n = n / 2;
            i++;
        }
        for (j = i - 1; j >= 0; j--)
            printf("%d", a[j]);
        printf("\n");
    }
}

 

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2052Picture

Problem Description Give you the width and height of the rectangle,darw it.

 

Input Input contains a number of test cases.For each case ,there are two numbers n and m (0 < n,m < 75)indicate the width and height of the rectangle.Iuput ends of EOF.   Output For each case,you should draw a rectangle with the width and height giving in the input.
after each case, you should a blank line.   Sample Input 3 2

 

Sample Output +---+ |   | |   | +---+

 

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int w, h;
    int i, j;
    while (scanf("%d %d", &w, &h) != EOF)
    {
        cout << "+";
        for (i = 0; i < w; i++)
            printf("-");
        cout << "+" << endl;

        for (i = 0; i < h; i++)
        {
            cout << "|";
            for (j = 0; j < w; j++)
                printf(" ");
            cout << "|" << endl;
        }

        cout << "+";
        for (i = 0; i < w; i++)
            printf("-");
        cout << "+" << endl;
    }
}

 

 

 

 

 

 

 

 

 

 

标签：2051,case,int,number,base,each,2060,Input,HDOJ
From： https://www.cnblogs.com/elegantcloud/p/17176679.html