哦豁，来看计算方式了吧👀️

解方程问题。

（v1 - v2) * a = n

(v1 + v2) * b = n;

解出v1, v2 输出即可。

#include <stdio.h>

int main()
{
	int n, a, b;
	scanf("%d %d %d", &n, &a, &b);
	int v1 ,v2; 
	v1 = (n / a + n / b) / 2;
	v2 = (n / b - n / a) / 2;
    printf("%d %d",v1, v2);
    return 0;
}

#include<iostream>
using namespace std;
int main(){
long long int  n,a,b;long long int v1, v2;
cin>>n>>a>>b;
v1=(n/a+n/b)/2.0;
v2= (n/b-n/a)/2.0;
cout<<v1<<' '<<v2;
    return 0;
}

#include<stdio.h>
int main()
{
	int n,a,b;
	




	
	scanf("%d%d%d",&n,&a,&b);



	int v1 = (n / a + n / b) / 2;




	
	int v2 = n / b - v1;
	v1 >= v2;
	
	
	
	
	
	printf("%d %d",v1, v2);
	
	
	
	
	return 0;
}

#include <stdio.h>

int main(void) {

int n, a, b;

int m = scanf("%d %d %d", &n, &a, &b);

long long m_1 = (long long)n * (a + b);

long long m_2 = (long long)n * (a - b);

long long mmm = 2LL * a * b;

int v_1 = m_1 / mmm;

int v_2 = m_2 / mmm;

printf("%d %d", v_1, v_2);

return 0;

}

#include<stdio.h> int main (){ int n,a,b; scanf("%d %d %d",&n,&a,&b); int v1=(n/a+n/b)/2;8 int v2=(n/b-n/a)/2; printf("%d %d",v1,v2); return 0; //不收徒，尤其115寝室和曹佳佳

#include <stdio.h>
int main()
{
    int a,b,c;
    int v1,v2,n1,n2;
    scanf("%d%d%d",&a,&b,&c);
    n1=a/b;
    n2=a/c;
    v2=(n2-n1)/2;
    v1=(n1+n2)/2;
    printf("%d %d",v1,v2);
    return 0;
}////B站搜索关注Siglota喵

跑道长是300

#include<stdio.h> int main() { int a,b,n; scanf("%d %d %d",&n,&a,&b); int j,y; n = (j-y)*a; n = (j+y)*b; printf("v1=%d v2=%d",j,y); return 0; } 为什么我这个输出的数据不正确

#include<stdio.h> int main() { int n,a,b; scanf("%d %d %d",&n,&a,&b); int v1,v2; v1=(n*(a+b))/(2ab); v2=(n*(a-b))/(2ab); printf("%d %d",v1,v2); return 0; }

#include <iostream>
using namespace std;
int main()
{
    int n,a,b,v1,v2;
    cin>>n>>a>>b;
    cout<<(n/b+n/a)/2.0<<' '<<(n/b-n/a)/2.0<<endl;
    return 0;
}

#include <iostream> using namespace std; int main() { int n,a,b; cin>>n>>a>>b; cout<<(an+nb)/(2ab)<<" "<<(an-nb)/(2ab)<<endl; return 0;

}