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SMPSUM - Iterated sums |
Please compute the sum of squares for the given numbers: a, a+1, ..., b-1, b.
Input
Two numbers: a and b separated by space, where 1 <= a <= b <=100.
Output
Computed sum: a*a + (a+1)*(a+1) + ... + (b-1)*(b-1) + b*b
Example
Input: 1 4 Output: 30
Example 2
Input: 5 6 Output: 61
| Added by: | kuszi |
| Date: | 2013-08-26 |
| Time limit: | 1s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 |
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2017-09-02 15:41:25
when a==b, it will simply output the square of either a or b. Last edit: 2017-09-02 16:04:53 |
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2017-06-20 22:36:14
what will happen if a == b? |
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2016-06-13 21:37:09
@ryukerg You are right, deriving a formula is a way. |
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2016-05-12 23:02:18
"Two numbers: a and b separated by SPACE" this is wrong. The two numbers are separated by "Enter". If you try to solve it when they are separated by space you will get a wrong answer as result. Edit: this comment is just valid if you use c++ and typeconversion from string to int! Last edit: 2016-05-12 23:44:19 |
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2016-04-13 21:39:38
This is possible to solve in O(1) time. Not just solvable in O(b-a+1) time; |
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2016-02-19 14:23:43 anonimowy
Simple |
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2016-01-06 15:32:33
its simple : since a<b for examble a=1 & b =4 it goes as (start with a) 1*1 + 2*2 + 3*3 + 4*4 (upto b) |
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2015-12-13 10:23:15
it says to compute the squares of all integers between a and b including both |
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2015-11-21 15:58:45 Mohit
Can someone please explain the question more clearly? What does a*a + (a+1)*(a+1) + ... + (b-1)*(b-1) + b*b mean? the ... part i.e upto what length this multiplication should go a*a + (a+1)*(a+1) only or upto a*a + (a+1)*(a+1)+ (a+2)*(a+2) or even more? |
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