这题其实很简单，我用的是最麻烦的方式，分4类：1，m,n都是奇数 2，m是奇数，n是偶数 3，m是偶数，n是奇数 4，m，n都是偶数

然后后面解析看代码

#include <bits/stdc++.h> using namespace std;

int jishu[310];

int main() { int m, n, sum = 0; cin >> m >> n; for(int i = 1; i <= 300; i++) { jishu[i] = 2 * i - 1; } //这一部分是预先分布，可以看作是一个300项的奇数数列，可以避免重复求和，后面只需引用里面的数即可 if(m % 2 == 1 && n % 2 == 1) for(int i = (m+1)/2; i <= (n+1)/2; i++) sum += jishu[i]; else if(m % 2 == 0 && n % 2 == 1) { int m1 = m + 1; for(int i = (m1 + 1)/2; i <= (n+1)/2; i++) sum += jishu[i]; } else if(m % 2 == 1 && n % 2 == 0) for(int i = (m + 1)/2; i <= n /2; i++) sum += jishu[i]; else if(m % 2 == 0 && n % 2 == 0) { int m1 = m + 1; for(int i = (m1 + 1)/2; i <= n/2; i++) sum += jishu[i]; } cout << sum << endl; return 0; }

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

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

cout << s;
return 0;

}

#include #include #include using namespace std; int main(){ int a,b; cin>>a>>b; int s=0; if(a%2==0) a+=1; for(int i=a;i<=b;i+=2) s=s+i;

cout<<s;

return 0;

}

#include #include #include using namespace std; int main(){ int a,b; cin>>a>>b; int s=0; if(a%2==0) a+=1;

for(int i=a;i<=b;i+=2) s=s+i;


cout<<s;

return 0;

}

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

#include<bits/stdc++.h>
using namespace std;

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

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

so easy

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