noip2002 string transformation

Use the folding method (adding the ASCII codes of the individual characters), but this is more likely to collide.

You can tie the hash function to the length, first and last parameters of the string, etc., so that it is more random.

Encyclopedia: Folding method.

Splitting a keyword into parts with the same number of digits (the last part can have different digits) and then taking the sum of these parts (rounding off) as the hash address, this method is called folding.

For example, every Western book has an international standard book number, which is a 10-digit decimal number, and if you want to use it as a keyword to build a hash table, when the library has less than 10,000 titles, you can use this method to construct a four-digit hash function.

Why do you want to go both ways? You can search directly.,Every time for all the strings that can be changed.,Change it.,If it's heavy, you can map or hash.。

Simple Deep Search. Pass 3 parameters. One passes the subscript of the currently selected number, one passes the sum of the selected number, and one passes the sum of the selected number.

#include

#include

using namespace std;

const int maxn = 22;

int a[maxn];

int ans,n,k;

bool isprim(int num)

void dfs(int cur,int cnt,int num)return ;

for(int i=cur;i<=n;i++)int main()

varmap:array[-1..9,-1..5] of boolean;The subscript of the second dimension of map is up to 5

num:array[-1..9,-1..5] of longint;The subscript of the second dimension of num is up to 5

i,j,n,m,x,y:longint;

beginfillchar(map,sizeof(map),true);

readln(n,m,x,y);

for i:=-1 to 9 do

beginmap[-1,i]:=false;The subscript for the second dimension of map needs to be 9

map[5,i]:=false;

end;for j:=-1 to 5 do

beginmap[-1,j]:=false;

map[9,j]:=false;

end;map[x+2,y+1]:=false;

map[x+1,y+2]:=false;

map[x-1,y+1]:=false;

map[x-2,y+1]:=false;

map[x-2,y-1]:=false;

map[x-1,y-1]:=false;

map[x+1,y-2]:=false;

map[x+2,y-1]:=false;

num[0,0]:=1;

for i:=0 to n do

for j:=0 to m do

Beginif map[i,j]=false then num[i,j]:=0 The subscript of the first dimension of the Else Map is taken to n, and the subscript of the second dimension is taken to m

beginnum[i,j]:=num[i-1,j]+num[i,j-1];The subscript of the first dimension of num is taken to n, and the subscript of the second dimension is taken to m

end;if (i=1) and (j=0) then num[i,j]:=1;

end;writeln(num[i,j]);

end.

……map:array[-1..9,-1..5] of boolean;

You should be fine if you change -1 to -2.

This question cannot be arranged quickly, and using it will shuffle the order and lead to errors. Examples:

10,8,17,5 is 2 times, 5 8 10 17 is 3 times, you kind of like a merge fruit.