纯水题

for 循环直接解决。
代码：

#include<iostream>
#include<cmath>
#define QwQ return 0;
using namespace std;
int n,m;
int ans;
int main(){
	cin>>n>>m;
	for(int i=n;i<=m;i++){
		if(i&1)ans+=i;
	}
	cout<<ans;
	QwQ
}

这一题十分简单

//循环累加就行了
//时间复杂度为指数级
#include <iostream>
#include <algorithm>
using namespace std;
int n, m, i;int main() {
	int sum = 0;
	cin >> n >> m;
	for (i = n;i <= m;i++)
		if (i % 2 == 1)
			sum += i;
	cout << sum << endl;
}

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

直接数学等差数列公式走起！

#include<bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f
using namespace std;
int n,m;
signed main(){
	cin>>n>>m;
	if(!(n%2))n++;
	if(!(m%2))m--;
	cout<<(n+m)*(m-n+2)/4;
	return 0;
}

C++ :

#include<bits/stdc++.h>
using namespace std;
int n,m,d;
int main()
{
cin>>n>>m;
if(n%2==0)
	n=n+1;
for(int i=n;i<=m;i=i+2)
	{
	d=d+i;
	}
cout<<d;
}

Python :

# coding=utf-8
a,b=input().split()
c=int(a)
d=int(b)
if(c%2==0):
    c=c+1
if(d%2==0):
    d=d-1
e=(d-c)//2+1
print((c+d)*e//2)