[Analog] Hdu 5308 I wanna become A 24-point Master

Test instructions

Give you n number, each number is N,

Then the n number is subtraction, each number must be used once, and only once, and the number of computations will have to be used once.

That is to do n-1 times, ask can get 24, output process XI? XJ number I and the number of J do? Operation this form

Ideas:

Because N is likely to be large, it certainly requires a perfect structure.

The simplest way is to construct a 4*6

Constructs 4 needs 5 numbers, constructs 6 needs 7 numbers

That means at least 12 numbers, the remaining number is subtracted to 0, and then the multiplication is good.

But it's important to note that 13 doesn't satisfy this rule.

So 1~13 special sentence, more than 14 of the output by law is good ~

Code:

#include "cstdlib" #include "Cstdio" #include "CString" #include "Cmath" #include "queue" #include "algorithm" #include "    Iostream "using namespace Std;int main () {int n;            while (scanf ("%d", &n)!=-1) {if (n<=3) {puts ("-1");        Continue            } if (n==4) {puts ("1 * 2");            Puts ("5 + 3");            Puts ("6 + 4");        Continue            } if (n==5) {puts ("1 * 2");            Puts ("6 * 3");            Puts ("7-4");            Puts ("8/5");        Continue            } if (n==6) {puts ("1 + 2");            Puts ("3 + 4");            Puts ("5-6");            Puts ("7 + 8");            Puts ("9 + 10");        Continue            } if (n==7) {puts ("1 + 2");            Puts ("8 + 3");            Puts ("9/4");            Puts ("5/6");            Puts ("7 + 11");            Puts ("12 * 10");        Continue      } if (n==8) {      Puts ("1 + 2");            Puts ("9 + 3");            Puts ("5-4");            Puts ("6 * 11");            Puts ("7 * 12");            Puts ("8 * 13");            Puts ("10 + 14");        Continue  } if (n==9) {puts ("1/2"); 1 puts ("3-10");  8 Puts ("4 + 5"); Puts ("12 + 6"); Puts ("13/7");  3 puts ("One by One *"),//15 puts ("8-9");            0 puts ("15 + 16");        Continue  } if (n==10) {puts ("1 + 2");  Puts ("3 + 4");  Puts ("5 + 6");  Puts ("7-8"); 0 puts ("9 * 14");            0 puts ("+ +"),//16 puts ("16/10"),//17 4 puts ("17 + 11");//18 24            Puts ("18 + 15");        Continue  } if (n==11) {puts ("1 + 2");  Puts ("3 + 4");    13 22        Puts ("13/5");  2 puts ("+ +"),//15 ("6-7") puts; 0 puts ("8 * 16"); 0 puts ("9 * 17"); 0 puts ("Ten *"),//19 0 puts ("one *"),//20 0 puts ("+ +)",//21 C        Ontinue; } if (n==12) {puts ("1 + 2");//13 2n printf ("%d + 3\n", n+1);//14 3n printf ("%d + 4\n", n+2); 4n printf ("%d + 5\n", n+3); 5n printf ("%d + 6\n", n+4); 6n printf ("%d/7\n", n+5); (n+6) 6 puts ("8 + 9"); 2n printf ("%d + 10\n", n+7); 3n printf ("%d + 11\n", n+8); 4n printf ("%d/12\n", n+9); (N+10) 4 printf ("%d *%d\n", n+6,n+10);        (N+11) Continue;  } if (n==13) {puts ("1/2"); 1 puts ("3-14");  Puts ("4 + 5"); Puts ("16/6"); 17 2           Puts ("//18"); puts ("7-8"); 0 puts ("9 * 19"); 0 puts ("10 * 20"); 0 puts ("11 * 21"); 0 puts ("12 * 22"); 0 puts ("13 * 23"); 0 puts ("24 + 18");        Continue; } puts ("1 + 2"); 2n printf ("%d + 3\n", n+1); 3n printf ("%d + 4\n", n+2); 4n printf ("%d + 5\n", n+3); 5n printf ("%d + 6\n", n+4); 6n printf ("%d/7\n", n+5); (n+6) 6 puts ("8 + 9"); 2n printf ("%d + 10\n", n+7); 3n printf ("%d + 11\n", n+8); 4n printf ("%d/12\n", n+9); (N+10) 4 printf ("%d *%d\n", n+6,n+10);        (n+11) puts ("13-14");//n+12 0 int tep=n+12;            for (int i=15;i<=n;i++) {printf ("%d *%d\n", i,tep);        tep++;    } printf ("%d +%d\n", n+11,tep); } return 0;}

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[Analog] Hdu 5308 I wanna become A 24-point Master