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找整除

[编程题]找整除

热度指数：3552 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

牛牛想在[a, b]区间内找到一些数满足可以被一个整数c整除,现在你需要帮助牛牛统计区间内一共有多少个这样的数满足条件？

输入描述:

首先输入两个整数a,b,（-5*10^8 ≤ a ≤ b ≤ 5*10^8) 接着是一个正整数c（1 <= c <= 1000）

输出描述:

输出一个整数表示个数。

示例1

输入

0 14 5

输出

算法知识视频讲解

牛客338808号

  importjava.util.Scanner; 

  /** 

   * Created by Administrator on 2017/6/12. 

   */ 

  publicclassMain { 

      publicstaticvoidmain(String[] args) { 

          Scanner sc = newScanner(System.in); 

          String arr[] = sc.nextLine().split(" "); 

          inta = Integer.parseInt(arr[0]); 

          intb = Integer.parseInt(arr[1]); 

          intc = Integer.parseInt(arr[2]); 

          intcount = 0; 

          for(;a <= b;a++){ 

              if(a % c == 0) { 

                  count = 1; 

                  break; 

              } 

          } 

          System.out.println(count + (b-a)/c); 

      } 

  }

发表于 2017-06-13 10:45:36 回复(0)

Python

Mrzhao201711212243806

python3--这种题，只想说呵呵，有意思么
a,b,c=map(int,input().split())
if a%c==0:
    print((b-a)//c+1)
else:
    print((b-(a//c+1)*c)//c+1)

发表于 2017-12-06 23:51:11 回复(0)

gwhmma

#include <iostream>
//#include <string>
//#include <vector>
using namespace std;

int Countnum(int a, int b, int c)
{
    int ans=0;
    if(a%c==0)
        ans++;
    if(b%c==0)
        ans++;
    a++;
    b--;
    if(b%c==0)
        ans++;
    int tmp=b-a;
    ans+=tmp/c;
    return ans;
}

int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    cout<<Countnum(a,b,c);
    return 0;
}

发表于 2017-03-31 20:54:07 回复(1)

魅影骑士


	importjava.util.Scanner;



	//在[a, b]区间内找到一些数满足可以被一个整数c整除,



	publicclassMain {



	    publicstaticvoidmain(String[] args){



	        Scanner in = newScanner(System.in);



	        while(in.hasNext()){



	            intbegin = in.nextInt();



	            intend = in.nextInt();



	            intnum = in.nextInt();



	            //int count = 0;



	            //for(int i = begin; i <= end; i++){



	            //  if(i % num == 0){



	                //  count++;



	                //}



	            //}



	            //System.out.println(count);



	            while(begin % num != 0){



	                begin++;



	            }



	            if(begin > end){



	                System.out.println(0);



	            }else{



	                System.out.println((end - begin) / num + 1);



	            }



	        }



	    }



	}

分析：调整a到一个可以被c整除的位置,然后以c为步长计算就好了

发表于 2017-03-24 16:10:57 回复(4)

C/C++

缘故

#include<iostream>
using namespace std;
int main(){
    int a,b,c;
    cin>>a>>b>>c;
    int t= b/c - a/c;
    if ( (a<=0&&b>=0)||(a%c==0&&a>=0)||(b%c==0&&b<=0) ) t++;
    cout<<t;
    return 0;      
}

编辑于 2017-07-25 17:12:58 回复(3)

nooBiee

import sys
 
try:
    while True:
        a,b,c = input().split()
        a = int(a)
        b = int(b)
        c = int(c)
        i0 = 0
        for i in range(a,b+1):
            if i%c == 0:
                i0 = i
                break
        n = 1 + (b-i0)//c
        print(n)
except:
    pass

发表于 2020-03-19 22:32:03 回复(0)

C/C++

长安云龙

#include <iostream>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    b -= b%c;
    cout<<((b - a)/c + 1);
    return 0;
}

b -= b%c 是要处理 -3 12 5 这种场景，不掐头去尾的话，会多算一个

发表于 2019-12-23 16:24:44 回复(0)

qinchuan

import java.util.Scanner;
 public class Main
 {
 public static void main(String [] args){
 Scanner scanner = new Scanner(System.in);
 int min=scanner.nextInt();
 int max=scanner.nextInt();
 int div=scanner.nextInt();
 int count=0;
 //for(int i=min;i<=max;i++){
 //    if(i%div==0){
 //        count=(max-i)/div+1;
 //       break;
 //   }
 //}
 int b=min;
 while(b<=max){
 if(b%div==0){
 count=(max-b)/div+1;
 break;
 }
 b++;
 }
 System.out.println(count);
 scanner.close();
 }
 }

编辑于 2018-12-04 18:17:49 回复(0)

进步不断

  a,b,c=map(int,input().split()) 
  suml,flag=0,0 
  ford inrange(a,b+1): 
      ifd%c==0: 
          flag=d 
          break 
  print(1+(b-d)//c) 

发表于 2018-07-19 00:51:38 回复(0)

奔跑的飛魚

#include<iostream>
using namespace std;

int find(int a, int b, int c)
{  if (b <= 0)  return find(-1 * b, -1 * a, c);  else  {  if (a < 0)  return find(0, b, c) + find(1, -1 * a, c);  else  {  int result = 0;  if (a == 0 || (a%c == 0 && b%c == 0))  result++;  if (c > b)  return result;  int tmp1 = a / c;  int tmp2 = b / c;  result += tmp2 - tmp1;  return result;  }  }
}

int main()
{  int a, b, c;  cin >> a >> b >> c;  cout << find(a, b, c);
    return 0;
}

编辑于 2018-06-08 19:53:11 回复(0)

edithw

//没有循环的方法，通过所有测试用例
//如果区间包含0需要单独加1
#include<iostream>
using namespace std;
int main(){
    int a, b, c;
    cin >> a >> b >> c;
    int count = b / c - a / c;
    if (b - a >= b) count++;
    cout << count << endl;
}

发表于 2017-11-03 10:22:01 回复(0)

梦向南

  //找到第一个，后面的也就都确定了，不用一直循环下去 

  #include<iostream> 

  using namespace std; 

  intmain() 

  { 

      inta,b,c; 

      while(cin>>a>>b>>c) 

      { 

          intcont=0; 

          for(inti=a;i<=b;i++) 

          { 

              if(i%c==0) 

              { 

                  cont=(b-i)/c + 1; 

                  break; 

              } 

          } 

          cout<<cont<<endl; 

      } 

      return0; 

  }

发表于 2017-07-01 23:52:23 回复(0)

向日葵葵

#include<iostream>

int main()

{

using namespace std;

int a, b;

cout << "Enter a section" << endl;

cin >> a >> b;

int c;

cout << "Enter a divisor" << endl;

cin >> c;

int count = 0;

for (int i = a; i <= b; i++)

{

if ((i%c) == 0)

count++;

}

cout << count << " numbers can be divided.";

system("pause");

发表于 2017-06-13 20:02:31 回复(0)

Jimmy_gg

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>

int main(){
	using namespace std;
	int a, b, c;
	while(cin >> a >> b >> c){
		int ans = 0;
		int first;
		if(a % c)
			first = c - a % c + a;
		else 
			first = a;
		for(int i = first; i <= b; i = i + c)
			ans ++;
		cout << ans << endl;
	}
	return 0;
}

发表于 2017-04-14 11:26:28 回复(0)

无瑕2017

#-*-coding:utf-8-*-
arr = map(int, raw_input().split(' '))
n = 0
for m in range(arr[0], (arr[1]+1)):
    if arr[2] != 0 and m % arr[2] == 0:
        n = n + 1
print n

您的代码已保存
内存超限:您的程序使用了超过限制的内存
case通过率为50.00%

本地测试妥妥的啊

编辑于 2017-04-11 11:30:38 回复(2)

trtrtr

  importjava.util.*; 
    
  publicclassMain{ 

      publicstaticvoidmain(String[] args){ 
          Scanner sc
  =newScanner(System.in); 
          longa = sc.nextInt(); 
          longb = sc.nextInt(); 
          longc = sc.nextInt(); 
          longres;
                  //判断ab是否同号，分别解答

          if(a*b<=0) res =
  Math.abs(a)/c+Math.abs(b)/c+1; 
          elseres = b/c-a/c; 
          System.out.print(res); 
      } 

   
  

发表于 2017-04-09 01:40:07 回复(0)

coco112

  1、找到左边的第一个可以整除的数
    2、找到右边的第一个可以整除的数
     3、sum = （右边的数减去左边的数）/c +1

        

      
 intmain()  
  { 
      inta,b,c; 

      cin>>a>>b>>c; 
      intt1=0; 
      for(inti=a;i<=b;i++) 
      { 
          if(i%c ==0) 
          { 
              t1 = i/c; 
              break; 
          } 
      } 
      intt2=0; 
      for(inti = b;i>=a;i--) 
      { 
          if(i%c ==0) 
          { 
              t2 = i/c; 
              break; 
          } 
      } 
    

      cout<<(t2-t1)+1<<endl; 
        
      return0; 
  } 

发表于 2017-04-08 19:42:06 回复(0)

wanlanwalan

#include<iostream>
using namespace std;
void getExactDivision(long long int a, long long int b, long int c)
{
 int num_an = 0;
 int num_bn = 0;
 num_bn = b / c;//求出倍数    
 num_an = a / c;
 if (a != 0)//判断是否为0    
 {
  if (b / a < 0)//说明 异号        
  {
   cout << (num_bn - num_an + 1) << endl;
  }
  else
   cout << (num_bn - num_an) << endl;;
 }
 else if (a == 0)
 {//等于0        
  cout << (num_bn - num_an + 1) << endl;
 }
}

 int main()
{
 long long int min_an, max_bn;
 cin >> min_an >> max_bn;//输入范围    
 long int num_cn;
 cin >> num_cn;
 //输入c    
 getExactDivision(min_an, max_bn, num_cn);
 return 0;
}

发表于 2017-04-05 11:55:39 回复(0)

笑忘书生

#include <iostream>
using namespace std;

int main()
{
	int n1, n2,n3;
	int count = 0;
	cin >> n1 >> n2 >> n3;
	if (n1 < 0 && n2 < 0)
		count = n2 / n3 - n1 / n3;
	if (n1 <= 0&&n2>=0)
		count = n2 / n3 - n1 / n3 + 1;
	if (n1 > 0 && n2 > 0)
		count = n2 / n3 - n1 / n3;
	cout << count;
	return 0;
}

发表于 2017-03-27 17:32:26 回复(0)

bian6


	#include "stdio.h"



	intmain()



	{



	   inta,b,c;



	   while(scanf("%d%d%d",&a,&b,&c)!=EOF)



	   {



	      intnum1,num2,num;



	      if(a<=0)



	      {



	          num1=(-a/c);



	          num2=b/c;



	          num=num1+num2+1;



	      }



	      else



	      {



	          num1=a/c;



	          num2=b/c;



	          num=num2-num1;



	      }



	       printf("%d\n",num);



	       



	   }



	}