l***i
发帖数: 1309 | 1 two integer array:
a1, a2, ..., an
b1, b2, ..., bn
Given a1 + a2 + ... + an = b1 + b2 + ... + bn
and a1^2 + a2^2 + ... + an^2 = b1^2 + b2^2 + ... + bn
and a1 <= a2 <= ... <= an
b1 <= b2 <= ... <= bn
Is it true that
a1 = b1, a2 = b2, ... an = bn? | h***t
发帖数: 2540 | | s****r
发帖数: 11 | 3 Can you elaborate a little more? Thanks.
【在 h***t 的大作中提到】
: NO
| l***i
发帖数: 1309 | 4 If you said no, give an example or an existence proof.
If yes, say something about proof ideas. | p*********9
发帖数: 30 | 5 Yes, a1=b1.
suppose a1+t=b1 where t>0
=> a2-t=b2, with a1^2 + a2^2 = b1^2 + b2^2
=> t=a2-a1
with the assumption a1+t=b1 => a2=b1 => a1=b2
so it's conflict with a1
done!
【在 l***i 的大作中提到】
: two integer array:
: a1, a2, ..., an
: b1, b2, ..., bn
: Given a1 + a2 + ... + an = b1 + b2 + ... + bn
: and a1^2 + a2^2 + ... + an^2 = b1^2 + b2^2 + ... + bn
: and a1 <= a2 <= ... <= an
: b1 <= b2 <= ... <= bn
: Is it true that
: a1 = b1, a2 = b2, ... an = bn?
| r*****r
发帖数: 630 | 6 the last term on the right hand side of the second equation is bn^2 or bn?
【在 l***i 的大作中提到】
: two integer array:
: a1, a2, ..., an
: b1, b2, ..., bn
: Given a1 + a2 + ... + an = b1 + b2 + ... + bn
: and a1^2 + a2^2 + ... + an^2 = b1^2 + b2^2 + ... + bn
: and a1 <= a2 <= ... <= an
: b1 <= b2 <= ... <= bn
: Is it true that
: a1 = b1, a2 = b2, ... an = bn?
| s*x
发帖数: 3328 | 7 one counter-example, for example:
0^2+3^2+5^2+6^2+9^2+10^2+12^2+15^2
=620
=1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2
【在 l***i 的大作中提到】
: two integer array:
: a1, a2, ..., an
: b1, b2, ..., bn
: Given a1 + a2 + ... + an = b1 + b2 + ... + bn
: and a1^2 + a2^2 + ... + an^2 = b1^2 + b2^2 + ... + bn
: and a1 <= a2 <= ... <= an
: b1 <= b2 <= ... <= bn
: Is it true that
: a1 = b1, a2 = b2, ... an = bn?
| p*****n
发帖数: 368 | 8 so easy ah,
3^2+4^2=0^2+5^2
【在 l***i 的大作中提到】
: If you said no, give an example or an existence proof.
: If yes, say something about proof ideas.
| p*****n
发帖数: 368 | 9 ...
【在 p*********9 的大作中提到】
: Yes, a1=b1.
: suppose a1+t=b1 where t>0
: => a2-t=b2, with a1^2 + a2^2 = b1^2 + b2^2
: => t=a2-a1
: with the assumption a1+t=b1 => a2=b1 => a1=b2
: so it's conflict with a1: done!
| p*********9
发帖数: 30 | 10 but 3+4 != 0+5
【在 p*****n 的大作中提到】
: so easy ah,
: 3^2+4^2=0^2+5^2
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