sum of consecutive positive integers

[shaveta]

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

[rgb]

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]

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]

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

[pixcee2000]

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

[sumanjha]

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 ;
}

[sumanjha]

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 ;
}