At first, I thought of going somewhere else, but the time difference between the two places is more than 12 hours.
The result will be the same, but due to precision issues, I referred to a non-precision solution on the discuss page.
First, the difference between the two watches per second is y=(abs(n-m)/86400).
And how many times do we need to have the exact 12-hour difference?
t=(43200/y)
So the first watch is delayed by (86400-n)/86400 per second. This should not be called a delay, but a comparison with the actual number of seconds.
And our t is calculated based on the delay per second.
So t*(86400-n)/86400 is the time compared to the actual number of seconds for the first watch.
After that, we divide it by 60 to get the minutes.
Remember to use rounding to get the minutes and hours.
//
// GGGGGGGGGGGGG CCCCCCCCCCCCC AAA
// GGG::::::::::::G CCC::::::::::::C A:::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::::A
// G:::::G GGGGGG C:::::C CCCCCC A:::::::::A
//G:::::G C:::::C A:::::A:::::A
//G:::::G C:::::C A:::::A A:::::A
//G:::::G GGGGGGGGGGC:::::C A:::::A A:::::A
//G:::::G G::::::::GC:::::C A:::::A A:::::A
//G:::::G GGGGG::::GC:::::C A:::::AAAAAAAAA:::::A
//G:::::G G::::GC:::::C A:::::::::::::::::::::A
// G:::::G G::::G C:::::C CCCCCC A:::::AAAAAAAAAAAAA:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::A A:::::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A A:::::A
// GGG::::::GGG:::G CCC::::::::::::C A:::::A A:::::A
// GGGGGG GGGG CCCCCCCCCCCCCAAAAAAA AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 30005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
#define X first
#define Y second
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int n,m;
void solve(){
Int d=abs(n-m);
Int sec=43200*(86400-n)/d;
Int min=43200*(86400-n)/(d*60);
Int a=43200*(86400-n),b=(d*60);
if((a%b)*2>=b){
min++;
}
Int h=(min/60)%12,mm=min%60;
if(h==0)h=12;
printf("%lld %lld %02lld:%02lld\\n",n,m,h,mm);
}
int main() {
ios_base::sync_with_stdio(0);
while(~scanf("%lld%lld",&n,&m)){
solve();
}
}