/* James opened a candy store. He was ingenious: He made water fructose into four one package and seven one package. Candy cannot be sold in a package.

When a child comes to buy sugar, he uses these two packages for combination. Of course, the quantity of some candy cannot be combined, for example, to buy 10 sugar.

You can test it on a computer. In this case, the maximum number of packages that cannot be purchased is 17. Any number greater than 17 can be combined with 4 and 7.

The requirement for this question is to find the maximum number that cannot be combined when the number of two packages is known.

Input:

Two positive integers, indicating the number of sugar in each package (no more than 1000)

Required output:

A positive integer that indicates the maximum number of sugar that cannot be bought

For example:

User input:

4 7

The program should output:

17

For example:

User input:

3 5

The program should output:

7 */

Method 1:

[Cpp]

# Include "stdio. h"

# Include "stdlib. h"

# Include "time. h"

# Deprecision max 100000000

Int main ()

{Int m, n;

Long int start, finish;

Scanf ("% d", & m, & n );

Start = clock ();

Long int * p;

P = (long *) malloc (sizeof (long) * max); // open up space and create a table 0, 1

P [0] = 1; // Initialization

For (long int I = 1; I <max; I ++)

{P [I] = 0; // Initialization

If (I> = m & p [I-m]) p [I] = 1;

Else

If (I> = n & p [I-n]) p [I] = 1;

}

Long int kk = 0;

For (long int I = 0; I <max; I ++)

{If (I> kk &&! P [I])

Kk = I;

}

Printf ("% ld", kk );

Finish = clock ();

Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );

Printf ("\ n ");

System ("pause ");

}

# Include "stdio. h"

# Include "stdlib. h"

# Include "time. h"

# Deprecision max 100000000

Int main ()

{Int m, n;

Long int start, finish;

Scanf ("% d", & m, & n );

Start = clock ();

Long int * p;

P = (long *) malloc (sizeof (long) * max); // open up space and create a table 0, 1

P [0] = 1; // Initialization

For (long int I = 1; I <max; I ++)

{P [I] = 0; // Initialization

If (I> = m & p [I-m]) p [I] = 1;

Else

If (I> = n & p [I-n]) p [I] = 1;

}

Long int kk = 0;

For (long int I = 0; I <max; I ++)

{If (I> kk &&! P [I])

Kk = I;

}

Printf ("% ld", kk );

Finish = clock ();

Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );

Printf ("\ n ");

System ("pause ");

} // A small number of optimizations. If n consecutive p [I] = 1 occurs when the number is larger, then no judgment is needed for all the subsequent numbers. So optimized, n <m

Method 2:

[Cpp]

# Include "stdio. h"

# Include "stdlib. h"

# Include "time. h"

# Deprecision max 100000000

Int main ()

{Int m, n, temp;

Long int start, finish;

Scanf ("% d", & m, & n );

Start = clock ();

Long int * p;

P = (long *) malloc (sizeof (long) * max); // open up space and create a table 0, 1

P [0] = 1; // Initialization

If (m <n)

{Temp = m; m = n; n = temp;} // ensure n <m

Temp = 0; // Initialization

For (long int I = 1; I <max; I ++)

{P [I] = 0; // Initialization

If (I> = m & p [I-m]) p [I] = 1;

Else

If (I> = n & p [I-n]) p [I] = 1;

If (p [I]) temp ++; else temp = 0;

If (temp = n) {printf ("% ld", I-n); break ;}

}

/* Long int kk = 0;

For (long int I = 0; I <max; I ++)

{If (I> kk &&! P [I])

Kk = I;

}

Printf ("% ld", kk );*/

Finish = clock ();

Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );

Printf ("\ n ");

System ("pause ");

}

# Include "stdio. h"

# Include "stdlib. h"

# Include "time. h"

# Deprecision max 100000000

Int main ()

{Int m, n, temp;

Long int start, finish;

Scanf ("% d", & m, & n );

Start = clock ();

Long int * p;

P = (long *) malloc (sizeof (long) * max); // open up space and create a table 0, 1

P [0] = 1; // Initialization

If (m <n)

{Temp = m; m = n; n = temp;} // ensure n <m

Temp = 0; // Initialization

For (long int I = 1; I <max; I ++)

{P [I] = 0; // Initialization

If (I> = m & p [I-m]) p [I] = 1;

Else

If (I> = n & p [I-n]) p [I] = 1;

If (p [I]) temp ++; else temp = 0;

If (temp = n) {printf ("% ld", I-n); break ;}

}

/* Long int kk = 0;

For (long int I = 0; I <max; I ++)

{If (I> kk &&! P [I])

Kk = I;

}

Printf ("% ld", kk );*/

Finish = clock ();

Printf ("\ nall time is: % lfs", (finish-start)/1000.0 );

Printf ("\ n ");

System ("pause ");

}