3 5 -> 4
5 3 -> 5
1 5 -> 0
5 1 -> 1
10 10 -> 18
99999 99999 -> 199996
2147483647 2147483647 -> 4294967292

I hope it helps!

answer = min(2*(n-1), 2*m-1);
//don't forget about 32-bit signed  integer overflow

import sys
n,m=map(int,input().split())
n1=n
m1=m
j=0
ans=0
if n1==1 or m1 ==1:
    print(0)
    sys.exit()
while n1>0 and m1>0:
    if ans%2==0 and j!=0:
        if m1>0:
            m1-=1
            ans+=1
            j+=1
        else:
            print(ans)
            sys.exit()
    elif ans%2==1 and j!=0:
        if n1>0:
            n1-=1
            ans+=1
            j+=1
    else:
        n1-=1
        m1-=1
        ans+=1
        j+=1
print(ans)

I print 2 * (n - 1) when n <= m and 2 * (m - 1) + 1 when n > m but got WA on test 13. What's wrong with my solution?

Oh, I forgot to delete my check for m = 1 (in this case I output 1). This test seem to be
1 1
0

Edited by author 03.10.2018 17:42

#include <stdio.h>
int main()
{
    int N, M;
    scanf("%d%d",&N,&M);
    long int res;
    if (N <= M)
        res = 2 * (N - 1);
    else
        res= 2 * (M - 1)+1;
    printf("\n%d",res);
    return 0;
}

think about :

F(N,M)  =  F(N-? , M- ? )

and give the initial values.

#include<stdio.h>
#include<string.h>
int f(int n,int m)
{
    if(n==1||m==1)
        return 0;
    if(n==2&&m>1)
        return 2;
    if(n==3&&m==2)
        return 3;
    if(m>2&&n>2)
    return f(n-2,m-2)+4;

}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    printf("%d",f(n,m));
    fflush(stdin);
    getchar();
    return 0;
}

Test 12 is overflow case. Be careful with data type.

hi,
in the acceptable answer the answer of 5 4 is 7 but you can easily understand that the real answer is 6(you can test it by drawing a simple 5 4 table).

here is a sample answer:

#include <iostream>

using namespace std;

int main() {
    int unsigned long long N , M , turn = 0;
 cin >> N >> M;



 while( N > 2 && M > 2){
     turn += 4;
     N -= 2;
     M -= 2;
 }
 if (N == 2 && M > 1 || M == 2 && N > 1)  turn += 2;

 if (M == 1 && N > 1) turn += 1;
  cout << turn << endl;

  system ("PAUSE");
  return 0 ;

}

System.out.println((Math.min(N - 1L, M - 1L) << 1) + (N > M ? 1 : 0));

I got WA 10, where is wrong?


///////////////////////
#include <iostream>
using namespace std;
int main()
{
    int N,M,sum,counter,mul;
    counter=0;
    cin>>N>>M;
    sum=M;
    mul=N*M;
    while(sum!=mul)
    {
        N--;
        sum+=N;
        counter++;
        if(sum==mul)
            break;
        M--;
        sum+=M;
        counter++;
    }
    cout<<counter<<endl;

}

input
2 1

output
0

this is correct. however this solution get WA. AC program writes 1, which is wrong in fact.

#include<cstdio>
#include<cstring>
#include<cassert>
#include<vector>
#include<list>
#include<queue>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<fstream>
#include<typeinfo>
#include<locale>
#include<iterator>
#include<valarray>
#include<complex>
using namespace std;

int main()
{
    int n,m;

    while(cin>>n>>m){
        int sum = 0,flag = 0,count = -1,x=n*m;
        n--;
        if(n==0 || m==0){
            cout<<"0"<<endl;
            continue;
        }
        while(sum < x){
            if(flag==0){
                sum += m;
                flag = 1;
                m--;
            }
            else if(flag==1){
                sum += n;
                flag = 0;
                n--;
            }
            count++;
        }
        cout<<count<<endl;
    }

    return 0;
}

here is my code, it has got verdict WA in test case 10....

program spiral;
var
 n,m,r:real;
begin
 readln(n,m);
 if m>=n then
  begin
   if n=1.0 then
    r:=0.0
   else
    if n=2.0 then
     r:=2.0
   else
    if frac(n/2)<>0 then
     r:=4.0*trunc(n/2)
   else
    r:=2.0*(n-1);
  end
 else
  begin
   if m=1.0 then
    r:=1.0
   else
    if m=2.0 then
     r:=3.0
   else
    if frac(m/2)<>0 then
     r:=4.0*trunc(m/2)+1.0
   else
    r:=2.0*m-1.0;
 end;
 writeln(r:0:0);
end.

I have correct answer on PC and wrong - here?Why?

What is it

I’ve AC, but I still have a question!

My first program used “unsigned long” type, and didn’t pass Test 12

Then I changed all types to “unsigned long long” , and program was AC.

So, my question is why my first program failed on Test12?

As you know n,m<=2^31-1;So, the biggest result is 0xFFFFFFFD! This result can be represented by unsigned long!

So, I didn’t announce my program, because it is bad style. I’m sure that those programmers who solved this issue can understand my question….

Best Wishes,
Veniamin.

#include <iostream>
using namespace std;
int main()
{
    unsigned long long n,m;
    cin>>n>>m;
    if(m>=n) cout<<2*(n-1)<<endl;
    if(n>m) cout <<2*(m-1)+1<<endl;
return 0;
}

#include<stdio.h>

int main(void){
    int n,m;
    unsigned int k;
    scanf("%d %d",&n,&m);
    k=0;
    while(n>2 && m>2){
        k=k+4;
        n-=2;
        m-=2;
    }
    if(n==2 && m>=2)
        k=k+2;
    else
        if(m==2 && n>2)
            k=k+3;
    else
        if(m==1 && n>1)
            k++;
    printf("%d",k);
return 0;
}