PUSSY_LICKING_LOLI_69's blog

Differences between (a+=b)%p; and a =(a+b)%p;
By PUSSY_LICKING_LOLI_69, history, 3 months ago, In English

I recently came across this problem while solving E1 of Codeforces Round 967 (Div. 2)

My original code used (a+=b)%p; and it worked perfectly with small test cases, but it gave WA when the test cases get larger. After i changed it to a =(a+b)%p; the code got accepted.

I'm a bit frustrated because it took me until after the contest was finished that i found the dumb mistake that probably stopped me from getting the Candidate Master title, so I made this post to discuss the differences between those 2 syntax to help people avoid the same mistake as me (and mostly just for me to vent T^T )

Here's my original code (Please don't pay attention to the dumb variable names hehe)

My Code

It only worked on test 1, and failed test 2. However, when i changed this line, it worked perfectly

//for(int j = 2;j<=k;j++) (dp2[i][j]+=dp2[i][j-1])%p;
for(int j = 2;j<=k;j++) dp2[i][j]=(dp2[i][j]+dp2[i][j-1])%p;

Can anyone explain why this single line broke my code? :(

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3 months ago, # |
← Rev. 2   Vote: I like it +25 Vote: I do not like it

Try for(int j = 2;j<=k;j++) (dp2[i][j]+=dp2[i][j-1])%=p;. Notice the equals sign after the mod ;)

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    3 months ago, # ^ |
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    Thanks it works now, but why does the one without the equal sign work with small numbers?

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      3 months ago, # ^ |
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      That's because the += operator goes, through, but the % operator doesn't do anything; its result is not assigned to anything.

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3 months ago, # |
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Just because.. this shouldn't take mod and it's pretty logical

The result of the % operation is not assigned to anything

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