题解 - 【入门】求数的量级？

0

hetao1401398 LV 6 @ 2024-6-10 14:13:08

直接if嵌套。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n/10000>=1&&n/10000<=9)
    {
        cout<<"wan";
    }
    else if(n/100000>=1&&n/100000<=9)
    {
        cout<<"shi wan";
    }
    else if(n/1000000>=1&&n/1000000<=9)
    {
        cout<<"bai wan";
    }
    else if(n/10000000>=1&&n/10000000<=9)
    {
        cout<<"qian wan";
    }
    else if(n/100000000>=1&&n/100000000<=9)
    {
        cout<<"yi";
    }
    else if(n/1000000000>=1&&n/1000000000<=9)
    {
        cout<<"shi yi";
    }
return 0;
}

先赞后看，养成习惯！

0

5256536 LV 6 @ 2024-6-10 13:03:57

函数

#include <bits/stdc++.h> 
using namespace std;
int n;
int number(int a) //判断位数
{
    int sum = 0;
    while (a>0)
    {
        a/=10;
        sum++;
    }
    return sum; //sum储存：这个数有几位
}
int main()
{
    cin >> n;
    if (number(n) == 5)
    {
        cout << "wan";
    }
    if (number(n) == 6)
    {
        cout << "shi wan";
    }
    if (number(n) == 7)
    {
        cout << "bai wan";
    }
    if (number(n) == 8)
    {
        cout << "qian wan";
    }
    if (number(n) == 9)
    {
        cout << "yi";
    }
    if (number(n) == 10)
    {
        cout << "shi yi";
    }
    return 0;
}

0

hetao16358938 LV 6 @ 2024-5-16 19:43:34

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n >= 10000 and n < 99999)
    {
        cout << "wan" << endl;
    }
    if(n >= 100000 and n < 999999)
    {
        cout << "shi wan" << endl;
    }
    if(n >= 1000000 and n < 9999999)
    {
        cout << "bai wan" << endl;
    }
    if(n >= 10000000 and n < 99999999)
    {
        cout << "qian wan" << endl;
    }
    if(n >= 100000000 and n < 999999999)
    {
        cout << "yi" << endl;
    }
    if(n >= 1000000000 and n < 9999999999)
    {
        cout << "shi yi" << endl;
    }
    return 0;
}

0

hetao9443956 LV 10 @ 2024-3-17 16:13:25

全网最简代码上代码！！！

#include <iostream>
using namespace std;
int main()
{
    string m;
    cin>>m;
    if (m.length()==5)
    {
        cout<<"wan";
    }
    else if (m.length()==6)
    {
        cout<<"shi wan";
    }
    else if (m.length()==7)
    {
        cout<<"bai wan";
    }
    else if (m.length()==8)
    {
        cout<<"qian wan";
    }
    else if (m.length()==9)
    {
        cout<<"yi";
    }
    else if (m.length()==10)
    {
        cout<<"shi yi";
    }
    return 0;
}

0

hetao3448344 LV 6 @ 2024-1-31 14:51:48

#include <iostream>//前方高能，前方高能。
using namespace std;
int num;
int main()
{
	int n;
    cin >> n;
    while (n>0)//注意这里是whlie循环。
    {
        n=n/10;
        num++;
    }
    if (num==5)//判断失控了！
    {
        cout << "wan";
    }
    else if (num==6)
    {
        cout << "shi wan";
    }
    else if (num==7)
    {
        cout << "bai wan";
    }
    else if (num==8)
    {
        cout << "qian wan";
    }
    else if (num==9)
    {
        cout << "yi";
    }
    else if (num==10)
    {
        cout << "shi yi";
    }
    return 0;//判断有那么亿点点长......
}//但是依然very easy。

0

hetao11426156 LV 9 @ 2024-1-24 21:28:19

很简单~~如果减去一个量级的数结果小于零，则此数低于此量级，用if-else if语句从低量级至高量级逐步查找即可(AC过的)

#include <iostream>
using  namespace std;
int main()
{
    int a;
    cin>>a;
    if (a-100000<0)
    {
        cout<<"wan";
    }
    else if (a-1000000<0)
    {
        cout<<"shi wan";
    }
    else if (a-10000000<0)
    {
        cout<<"bai wan";
    }
    else if (a-100000000<0)
    {
        cout<<"qian wan";
    }
    else if (a-1000000000<0)
    {
        cout<<"yi";
    }
    else if (a-10000000000<0)
    {
        cout<<"shi yi";
    }
    return 0;
}

0

hetao9499759 LV 8 @ 2023-11-19 14:56:09

超级简单现学现用点一个赞评论一下养成习惯

#include<bits/stdc++.h>
using namespace std;
char s[9999];
int main(){
scanf("%s",s+1);
int l=strlen(s+1);
switch(l){
case 5:
cout<<"wan";
break;
case 6:
cout<<"shi wan";
break;
case 7:
cout<<"bai wan";
break;		
case 8:
cout<<"qian wan";
break;
case 9:
cout<<"yi";
break;
case 10:
cout<<"shi yi";
break;
}
return 0;
}

0

hetao35628597 LV 6 @ 2023-8-31 14:19:03

ctrlC、ctrlV是真的香!AC通过！

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n>=10000)
    {
        if(n<=99999)
        {
            cout<<"wan";
        }
    }
    if(n>=100000)
    {
        if(n<=999999)
        {
            cout<<"shi wan";
        }
    }
    if(n>=1000000)
    {
        if(n<=9999999)
        {
            cout<<"bai wan";
        }
    }
    if(n>=10000000)
    {
        if(n<=99999999)
        {
            cout<<"qian wan";
        }
    }
    
    if(n>=1000000000)
    {
        if(n<=9999999999)
        {
            cout<<"shi yi";
        }
    }
    return 0;
}

0

hetao8483546 LV 6 @ 2023-8-30 9:43:34

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	if(n>10000 && n<99999){
		cout<<"wan";
	}
	else if(n>100000 && n<999999){
		cout<<"shi wan";
	}
	else if(n>1000000 && n<9999999){
		cout<<"bai wan";
	}
	else if(n>10000000 && n<99999999){
		cout<<"qian wan";
	}
	else if(n>100000000 && n<999999999){
		cout<<"yi";
	}
	else if(n>1000000000 && n<9999999999){
		cout<<"shi yi";
	}
	return 0;
}

0

hetao108158 LV 6 @ 2023-8-18 18:02:02

ctrl+cv什么意思？？

hetao3937345 @ 2023-8-26 17:29:00

复制粘贴

0

hetao1356087 LV 6 @ 2023-8-17 20:53:41

老方法还是最管用滴，（可借鉴，无需付赞）

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if (n < 100000)
    {
        cout << "wan" << endl;
    }
    else if (n < 1000000)
    {
        cout << "shi wan" << endl;
    }
    else if (n < 10000000)
    {
        cout << "bai wan" << endl;
    }
    else if (n < 100000000)
    {
        cout << "qian wan" << endl;
    }
    else if (n < 1000000000)
    {
        cout << "yi" << endl;
    }
    else
    {
        cout << "shi yi" << endl;
    }
    return 0;
}

0

hetao11402094 LV 7 @ 2023-8-16 17:16:32

#include <bits/stdc++.h>
using namespace std;
int main()
{
    unsigned long long n;//就算9999999999也可以
    cin >> n;
    //看他能整除谁
    if (n / 1000000000 > 0)
    {
        cout << "shi yi";
        return 0;
    }
    if (n / 100000000 > 0)
    {
        cout << "yi";
        return 0;
    }
    if (n / 10000000 > 0)
    {
        cout << "qian wan";
        return 0;
    }
    if (n / 1000000 > 0)
    {
        cout << "bai wan";
        return 0;
    }
    if (n / 100000 > 0)
    {
        cout << "shi wan";
        return 0;
    }
    if (n / 10000 > 0)
    {
        cout << "wan";
        return 0;
    }
    return 0;
}

0

hetao3021474 LV 7 @ 2023-8-16 14:04:40

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n >= 10000 and n < 99999)
    {
        cout << "wan" << endl;
    }
    if(n >= 100000 and n < 999999)
    {
        cout << "shi wan" << endl;
    }
    if(n >= 1000000 and n < 9999999)
    {
        cout << "bai wan" << endl;
    }
    if(n >= 10000000 and n < 99999999)
    {
        cout << "qian wan" << endl;
    }
    if(n >= 100000000 and n < 999999999)
    {
        cout << "yi" << endl;
    }
    if(n >= 1000000000 and n < 9999999999)
    {
        cout << "shi yi" << endl;
    }
    return 0;
}

这道题很简单，疯狂写if就行了。代码不易，谢谢点赞！😄 别忘了点❤️ 在复制

0

hetao5144529 LV 8 @ 2023-8-1 21:34:40

直接计算位数

#include <iostream>
using namespace std;
int main()
{
    long long n;
    string str_s,a[]={"wan","shi wan","bai wan","qian wan","yi","shi yi"};
    cin>>n;
    str_s=to_string(n);
    cout<<a[str_s.size()-5];
}

hetao390640 @ 2023-8-5 15:00:33

这个代码我没学过，但是我觉得你的网名都可以用来刷屏了。。。。。。

0

hetao7316863 LV 8 @ 2023-7-9 18:27:35

这种多条件的题可以用这种三角形的代码解决，具体原理看前两层左右就能看明白了

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int n;cin>>n;
    if (n>=10000){
        if(n>=100000){
            if(n>=1000000){
                if(n>=10000000){
                    if(n>=100000000){
                        if(n>=1000000000){
                            cout<<"shi yi";
                        }
                        else{cout<<"yi";}
                    }
                    else{cout<<"qian wan";}
                }
                else{cout<<"bai wan";}
            }
            else{cout<<"shi wan";}
        }
        else{cout<<"wan";}
    }
    return 0;
}

hetao390640 @ 2023-8-5 15:04:08

不像三角形，反而像个数字'5'

0

hetao363897 LV 6 @ 2023-3-14 22:09:51

#include <iostream>
using namespace std;
int main()
{
    long long n;
    cin >> n;
    if ((n/10000>0) && (n/10000<10))
    {
        cout << "wan";
    }
    if ((n/100000>0) && (n/100000<10))
    {
        cout << "shi wan";
    }
    if ((n/1000000>0) && (n/1000000<10))
    {
        cout << "bai wan";
    }
    if ((n/10000000>0) && (n/10000000<10))
    {
        cout << "qian wan";
    }
    if ((n/100000000>0) && (n/100000000<10))
    {
        cout << "yi";
    }
    if ((n/1000000000>0) && (n/1000000000<10))
    {
        cout << "shi yi";
    }
    return 0;
}

0

Maomao LV 6 @ 2022-9-17 22:54:06

这道题一如既往的简单

直接用字符串，数位问题。

数亿数有几位，（用 字符串变量名.length() 即可。），然后减四，得到另一个字符串数组（所有可能的结果）的索引值。

废话不多说，上代码！

#include<bits/stdc++.h>
using namespace std;
string x, y[7] = {"114514", "wan", "shi wan", "bai wan", "qian wan", "yi", "shi yi"}; 
int main()
{
    cin >> x;
    cout << y[x.length() - 4] << endl;
	re turn 0; 
}

小盆友们，您学会了吗？

复制？代码不易，先点赞，再复制吧！（不要忘了删倒数第二行的空格!)

0

hetao2343161 LV 7 @ 2022-9-16 21:22:57

#include <iostream>
using namespace std;
string a[12]={" "," "," "," ","wan","shi wan","bai wan","qian wan","yi","shi yi"};
int main()
{
    string s;
    cin>>s;
    int len=s.size();
    cout<<a[len-1];
    return 0;
}

0

hetao2343161 LV 7 @ 2022-9-16 21:22:29

#include <iostream>
using namespace std;
string a[12]={" "," "," "," ","wan","shi wan","bai wan","qian wan","yi","shi yi"};
int main()
{
    string s;
    cin>>s;
    int len=s.size();
    cout<<a[len-1];
    return 0;
}

0

hetao252162 LV 6 @ 2022-8-25 10:21:24

这段代码其实非常简单，只要疯狂地使用if语句就可以了。只要你有那么一点点耐心--（* @ *） --

#include <iostream> using namespace std; int main() { int n; cin >> n;//这是一个注释-- if (n / 10000 >= 1 and n / 10000 <= 9)//举个例子，10000（五位数中的最小数）除以10000 商为1。 99999（最大数）除以10000 商为9。下面同样道理。。 { cout << "wan"; } if (n / 100000 >= 1 and n / 100000 <= 9) { cout << "shi wan"; } if (n / 1000000 >= 1 and n / 1000000 <= 9) { cout << "bai wan"; } if (n / 10000000 >= 1 and n / 10000000 <= 9) { cout << "qian wan"; } if (n / 100000000 >= 1 and n / 100000000 <= 9) { cout << "yi"; } if (n / 1000000000 >= 1 and n / 1000000000 <= 9) { cout << "shi yi"; } return 0; }

52 条题解

很简单~~如果减去一个量级的数结果小于零，则此数低于此量级，用if-else if语句从低量级至高量级逐步查找即可(AC过的)

直接计算位数

【入门】求数的量级？

信息

状态

开发

支持

52 条题解

很简单~~如果减去一个量级的数结果小于零，则此数低于此量级，用if-else if语句从低量级至高量级逐步查找即可(AC过的)

直接计算位数

【入门】求数的量级？

信息

状态

开发

支持

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