都是乐色！！！ 全网最全解答！！！

#include <bits/stdc++.h>
using namespace std;
const int SB=1e6+10;
int main ()
{
    system("color 2");//别管
    int q,z;//创建
    cin>>q>>z;//输入
    int b=0;//存储变量
    for(;q<=z;q++){//循环判断
        if(q%2!=0){//判断单复
            b+=q;//若是单数，存起来
        }
    }
    cout<<b;//输出存储
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
//天下无双，个数平方
/*
这也钛坚氮了
*/
int enumk(int n,int m){
    if(m%2==1&&n%2==1)return (m-n)/2+1;
    if((m%2==1&&n%2==0)||(m%2==0&&n%2==1))return (m-n-1)/2+1;
    return (m-n-2)/2+1;
}
int main(){
    int x,y,n1,n2;
    cin>>x>>y;
    if(x==0)x=1;
    n1=enumk(1,y)*enumk(1,y);
    n2=enumk(1,x-1)*enumk(1,x-1);
    cout<<n1-n2;
}

大哥大姐！！救救题解吧！！！（跪）

一到很卷单的题。

AC题解：

#include<bits/stdc++.h>
using namespace std;
int main(){
	int m,n,ans=0;
    cin>>m>>n;
    for(int i=m;i<=n;i++){
        if(i%2!=0)ans+=i;
    }
    cout<<ans;
	return 0;
}

到底是谁不会这道题啊？？

原来是我啊

禁止发疯！！

#include using namespace std; int main() {

int m ,n , sum = 0;
cin >> m >> n ;
if( m % 2 == 0)
{
	m = m + 1;
}
for(int i = m; i <= n; i = i + 2)
{
	sum = sum + i;
}
cout<<sum;	
return 0;

}

#include<cstdio>
#include<cctype>
#include<string.h>
#include<math.h>
#include<cmath>
#include<algorithm>
using namespace std;
long long cnt=0;
int main()
{
	long long m,n;
	cin>>m>>n;
	for(int i=m;i<=n;i++) 
	{
		if(i%2!=0)
		{
			cnt+=i;
		}
	}
	cout<<cnt;
	return 0;
}

#include<iostream>
#include<iomanip>
using namespace std;
double n , x , sum ;
int main(){
	cin >> n;
	for ( int i = 1 ; i <= n ; i++ )
	{
		cin >> x ;
		sum += x ;
	}
	cout<<fixed<<setprecision(4)<<sum/n<<endl;
}

#include<iostream>
using namespace std;
int main()
{
	int m,n;
	cin>>m>>n;
	long long sum=0;
	if(m%2==0)//求>=m的最大奇数
	{
		m++;
	}
	for(int i=m;i<=n;i+=2)//直接从>=m的最大奇数开始+=2，sum+=i;
	{
		sum+=i;
	}
	cout<<sum;
	return 0;
}

#include<bits/stdc++.h> using namespace std; int m,n,ans; int main(){

cin >> m >> n;
if(m%2==0){
	m++;
}
for(int i=m;i<=n;i += 2)
{
	ans += i;
}
cout<<ans;
return 0;

}

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
	int n,m,a=0;
	cin>>n>>m;
	for(int i=n;i<=m;i++){
		if(i%2==1)a+=i;
	}
	printf("%d",a);
}

我们知道一个数$\ n$当它被按位与$1$时，会变成大于等于$n$，离$n$最近的奇数。 所以可得AC Code:

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int n, m;

    scanf("%d %d", &n, &m);

    n |= 1;  //变成 >=n的最近奇数
    int ans = 0;

    while (n <= m)
    {
        ans += n;
        n += 2;
    }
    cout << ans;
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main(void)
{

int qs1, qs2;
int all;

cin >> qs1 >> qs2;

int m = qs1 % 2;
int n = qs2 % 2;


if (qs1 != qs2)
{
	if (m == 0)

	{
		qs1 = qs1 + 1;
	}

	if (n == 0)
	{
		qs2 -= 1;
	}

	if (qs1 == qs2)
	{
		all = qs1;
	}
	else
	{
		for (qs1 = qs1, qs2 = qs2; qs1 < qs2; qs1 += 2)
		{
			all = all + qs1;
		}

		all += qs1;
	}
}
else if (qs1 == qs2)
{
	if (n == 0, m == 0)
	{
		all = 0;
	}

	if (n != 0, m != 0)
	{
		all = qs1 = qs2;
	}
}

cout << all;

return 0;
}

#include <stdio.h>

#include <math.h>

int main(void)

{

int qs1 = 0, qs2 = 0;

int all = 0;

scanf("%d%d", &qs1, &qs2);

int m = qs1 % 2;
int n = qs2 % 2;


if (qs1 != qs2)
{
	if (m == 0)
	{
		qs1 = qs1 + 1;
	}
	
	if (n == 0)
	{
		qs2 -= 1;
	}

	if (qs1 == qs2)
	{
		all = qs1;
	}
	else
	{
		for (qs1 = qs1, qs2 = qs2; qs1 < qs2; qs1 += 2)
		{
			all = all + qs1;
		}
	
		all += qs1;
	}
}
else if (qs1 == qs2)
{
	if (n == 0, m == 0)
	{
		all = 0;
	}
	
	if (n != 0, m != 0)
	{
		all = qs1 = qs2;
	}
}

printf("%d", all);

return 0;

}

#include<iostream>
using namespace std;
int n,m,sum;
int main()
{
cin>>n>>m;
if(n%2==0)
{
n+=1;
}
for(int i=n;i<=m;i+=1)
{
if(i%2==1)
{
sum=sum+i;
}
}
cout<<sum;
return 0;

}

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	double l,r;
    cin >> l >> r;
    int sum = 0;
    for (int i = l; i <=  r ;i++ )
    {
    	if(i%2 == 1)
    	{
    		sum += i;
    	}	
    }
  	cout <<sum <<endl;
}

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	double l,r;
	cin >> l >> r;
	int sum = 0;
	for(int i = l ; i <= r ; i++)
	{
		if(i%2 ==1)
		{
			sum +=i;
		}
	}
	cout << sum << endl;
}