hdu--5119--number of decision scenarios DP

In fact, the Beijing station DP are not expected to be difficult.

But.. ..

Dp[x,y] Indicates the number of scenarios with the first X number of XOR values ==y

When transferring, you can first assign the upper level fully to this layer and assume that a[i] does not participate in XOR

Then A[i] is different from the upper value or needs to traverse all the scenarios 2 times

I was worried about Tle, but N was only 40.

I didn't use the scroll array to open the 40 directly at first.

Then an open scrolling array I don't know why I've been re ...

A long time to find that I will dp[x,y] the first open to 3 can be opened into 2 No, I'm depressed.

Obviously only can take 0 or 1 ah because the result of num&1 is not only 0 or 1?

No words ...

Not to understand. If I understand, I'll make up the reason below.

1#include <iostream>2#include <cstring>3 using namespacestd;4 5typedefLong LongLL;6 intN;7 Const intSize =1000000;8LL dp[2][size+Ten];9 inta[ $];Ten  One voidSolve () A { -      for(inti =1; I<=n; i++ ) -     { the          for(intj =0; J<=size; J + + ) -         { -dp[i&1][J] = dp[! ( i&1)][j]; -         } +          for(intj =0; J<=size; J + + ) -         { +dp[i&1] [J^a[i]] + = dp[! ( i&1)][j]; A         } at     } - } -  - intMain () - { -Cin.sync_with_stdio (false); in     intT, M; - LL ans; toCIN >>T; +      for(intK =1; k<=t; k++ ) -     { theCIN >> N >>m; *Memset (DP,0,sizeof(DP)); $          for(inti =1; I<=n; i++ )Panax Notoginseng         { -CIN >>A[i]; the         } +dp[0][0] =1; AAns =0; the solve (); +          for(inti = m; I<=size; i++ ) -         { $Ans + = dp[n&1][i]; $         } -cout <<"Case #"<< k <<": "<< ans <<Endl; -     } the     return 0; -}

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1#include <iostream>2#include <cstring>3 using namespacestd;4 5typedefLong LongLL;6 intN;7 Const intSize =1000000;8LL dp[3][size+Ten];9 inta[ $];Ten  One voidSolve () A { -      for(inti =1; I<=n; i++ ) -     { the          for(intj =0; J<=size; J + + ) -         { -dp[i&1][J] = dp[! ( i&1)][j]; -         } +          for(intj =0; J<=size; J + + ) -         { +dp[i&1] [J^a[i]] + = dp[! ( i&1)][j]; A         } at     } - } -  - intMain () - { -Cin.sync_with_stdio (false); in     intT, M; - LL ans; toCIN >>T; +      for(intK =1; k<=t; k++ ) -     { theCIN >> N >>m; *Memset (DP,0,sizeof(DP)); $          for(inti =1; I<=n; i++ )Panax Notoginseng         { -CIN >>A[i]; the         } +dp[0][0] =1; AAns =0; the solve (); +          for(inti = m; I<=size; i++ ) -         { $Ans + = dp[n&1][i]; $         } -cout <<"Case #"<< k <<": "<< ans <<Endl; -     } the     return 0; -}

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Today

Nightmare

It's raining.

Waiting for a man's coffee

Waiting for a phone call from a person

hdu--5119--number of decision scenarios DP