ZOJ3870 Team Formation

1 /**2 Author:oliver3 problemid:zoj3870 Team Formation4 */5 /*6 Ideas7 1. Xor operation, use ^ will explode, think of binary;8 2. We can try to simulate a previous one ^ so long as the current bit result becomes larger;9 3. How do we use the binary system in general? Since it's going to explode, we'll save 1 of our position.Ten 4. How does the problem exist and how do I use it?  One 5. What I thought before was that every 1 of each number was saved, but that was not how to avoid the repetitive calculations. Take a look at someone else's blog and try to save up to 1 of the maximum number of bits per digit.  A 6. Think about why it is the highest position . - 7. Well, now is not to put all the high 1 are saved, and then we up (1,n) to the original and the remainder of the steps to change into and bin[] operation.  - What we want is to become bigger than max{a,b}. Then look at the bigger one, then we'll look for a larger number, and this time we're not counting and comparing, but comparing to the top of the rest.  the Note that we should be able to think of a violent way to go through the past, to do the optimization, that is, each time just add can and this team to complete the number of good.  - Note that this does not have duplicates (like A and b,b and a), not so. Because there is no discovery, each time is a traversal of this number is max{a,b}.  - Note that we are a one-man move backwards until 0->1.  - step can not write? The idea is very detailed.  + * - #include <cstdio> + #include <cstring> A #include <algorithm> at  - using namespace std; -  - const int MAXM = 100000+10; - int bin[100]; - int A[MAXM]; in int main () - { to int t,n; + scanf ("%d", &t); - While (t--) the { * memset (bin,0,sizeof bin); $ scanf ("%d", &n);Panax Notoginseng int X; - for (int i=0;i<n;i++) the { + scanf ("%d", &a[i]); A X=a[i]; the for (int j=31;j>=0;j--)//here if you start from 32, you try what happens. + if (x& (1<<j)) {//extract highest bit - bin[j]++; $ Break ; $ }  - }  - long long ans=0; the Int J; - for (int i=0;i<n;i++)Wuyi { the X=a[i]; - For (j=31;j>=0;j--) Wu if (x& (1<<J)) break; - For (; j>=0;j--)//bin[] Operations About if (! ( x& (1<<J))) { $ Ans+=bin[j]; - }  - } -  A printf ("%lld\n", ans); + }     the }

2015-04-29 22:30:05

ZOJ3870 Team Formation