pls help me to write a program such that we input an integer x,where x>0. For x, the program has to convert it into the sum of consecutive positive integers. for e.g. let x=10 output should be "10= 1+2+3+4" the total no. of consecutive positive integers in the sum expression should be maximal. for e.g. if x= 9 result should be 2+3+4 and not 4+5. If x=4 , the output should be "no answer"
thanx in advance
6 8933 rgb 37
New Member
one solution might be:
step 1: do a loop of 1+2+3+4+5+... until your SUM is greater than the INPUT and stop (or exit) the loop
step 2: do a loop of SUM-1-2-3-4-... until your VALUE is equal to INPUT (which is the answer) OR less than the INPUT (which is no answer)
note: you need to store 2 values, last added (i will call it y) and last subtracted+1 (i will call it x) for your final string answer doing x+(x+1)+(x+2)+. .. until you reach y.
rgb 37
New Member
correction on step 1: the SUM should be equal or greater than (not equal only) and if it's equal (you got the answer), else (SUM is greater than INPUT) do step 2.
D_C 293
Contributor
Interesting problem, and the solution is less than 30 lines of code. I would like to say that there will always be a trivial solution. N = sum(1,N) - sum(1,N-1). Surprisingly, the trend seems to be that this only happens when N is a power of 2. - 1 = 1
-
2 = 2
-
3 = 1+2
-
4 = 4
-
5 = 2+3
-
6 = 1+2+3
-
7 = 3+4
-
8 = 8
-
9 = 2+3+4
-
10 = 1+2+3+4
-
11 = 5+6
-
12 = 3+4+5
-
13 = 6+7
-
14 = 2+3+4+5
-
15 = 1+2+3+4+5
-
16 = 16
-
17 = 8+9
-
18 = 3+4+5+6
-
19 = 9+10
-
20 = 2+3+4+5+6
-
21 = 1+2+3+4+5+6
-
22 = 4+5+6+7
-
23 = 11+12
-
24 = 7+8+9
-
25 = 3+4+5+6+7
-
26 = 5+6+7+8
-
27 = 2+3+4+5+6+7
-
28 = 1+2+3+4+5+6+7
-
29 = 14+15
-
30 = 4+5+6+7+8
-
31 = 15+16
-
32 = 32
-
33 = 3+4+5+6+7+8
-
34 = 7+8+9+10
-
35 = 2+3+4+5+6+7+8
-
36 = 1+2+3+4+5+6+7+8
pls help me to write a program such that we input an integer x,where x>0. For x, the program has to convert it into the sum of consecutive positive integers. for e.g. let x=10 output should be "10= 1+2+3+4" the total no. of consecutive positive integers in the sum expression should be maximal. for e.g. if x= 9 result should be 2+3+4 and not 4+5. If x=4 , the output should be "no answer"
thanx in advance
pls help me to write a program such that we input an integer x,where x>0. For x, the program has to convert it into the sum of consecutive positive integers. for e.g. let x=10 output should be "10= 1+2+3+4" the total no. of consecutive positive integers in the sum expression should be maximal. for e.g. if x= 9 result should be 2+3+4 and not 4+5. If x=4 , the output should be "no answer"
thanx in advance
Question is really good ,One Soln is given below . Run and check with diff input.
#include<stdio. h>
int main()
{
int num,sum1=0,inde x=1,sum2=0;
printf("Enter a no(>0) ");
scanf("%d",&num );
if(num <= 0 )
{
printf("Wrong Input\n");
return 0;
}
for(index;;inde x++)
{
sum1=sum1+index ;
if(sum1>=num)
break;
}
if(sum1 == num )
{
if(num == index)
{
printf("NO ANSWER\n");
return 0 ;
}
printf("Num(%d) = ",num );
for(int temp=1;temp<=in dex;temp++)
printf(" %d +",temp);
printf("\b \n");
return 0;
}
for(int i=1;i<index;i++ )
{
sum2+=i;
if((sum1-sum2) == num )
{
if((i+1) == index)
{
printf("NO ANSWER\n");
return 0 ;
}
printf("Num(%d) = ",num );
for(int temp= i+1;temp<=index ;temp++)
printf(" %d +",temp);
printf("\b \n");
return 0 ;
}
}
printf("NO ANSWER\n");
return 0 ;
}
Question is really good ,One Soln is given below . Run and check with diff input.
#include<stdio. h>
int main()
{
int num,sum1=0,inde x=1,sum2=0;
printf("Enter a no(>0) ");
scanf("%d",&num );
if(num <= 0 )
{
printf("Wrong Input\n");
return 0;
}
for(index;;inde x++)
{
sum1=sum1+index ;
if(sum1>=num)
break;
}
if(sum1 == num )
{
if(num == index)
{
printf("NO ANSWER\n");
return 0 ;
}
printf("Num(%d) = ",num );
for(int temp=1;temp<=in dex;temp++)
printf(" %d +",temp);
printf("\b \n");
return 0;
}
for(int i=1;i<index;i++ )
{
sum2+=i;
if((sum1-sum2) == num )
{
if((i+1) == index)
{
printf("NO ANSWER\n");
return 0 ;
}
printf("Num(%d) = ",num );
for(int temp= i+1;temp<=index ;temp++)
printf(" %d +",temp);
printf("\b \n");
return 0 ;
}
}
printf("NO ANSWER\n");
return 0 ;
}
Hi All ,
Above Code will not work for some specific input.
odd number has two solution.
suppose num is 9 then "9=2+3+4" OR "9=4+5"
Run the following code and test it .
#include<stdio. h>
int main()
{
int num,sum1=0,inde x=1,sum2=0,Soln =0;
printf("Enter a no(>0) ");
scanf("%d",&num );
if(num <= 0 )
{
printf("Wrong Input\n");
return 0;
}
for(index;;inde x++)
{
sum1=sum1+index ;
if(sum1>=num)
break;
}
if(sum1 == num )
{
if(num == index)
{
printf("NO ANSWER\n");
return 0;
}
Soln++;
printf("Solutio n-%d : %d = ",Soln,num );
for(int temp=1;temp<=in dex;temp++)
printf(" %d +",temp);
printf("\b \n");
}
for(int i=1;i<index;i++ )
{
sum2+=i;
if((sum1-sum2) == num )
{
if((i+1) == index)
{
// printf("NO ANSWER\n");
break;
}
Soln++;
printf("Solutio n-%d : %d = ",Soln,num );
for(int temp= i+1;temp<=index ;temp++)
printf(" %d +",temp);
printf("\b \n");
break;
}
}
//if( Soln == 0 && num >= 9 && num/2 == (num-1)/2 ) //--> if U don't want more than one ,soln use this
if( num >= 9 && num/2 == (num-1)/2 ) //--> if U want more than one soln, use this
{
Soln++;
printf("Solutio n-%d : %d = %d + %d\n",Soln,num ,num/2,num/2+1);
return 0;
}
if(Soln==0)
printf("NO SOLUTION\n");
return 0 ;
}
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