Usaco Chapter 1 Section 5

1. Number triangle, poj1163, not to mention.

Code:

# Include <cstdio>
# Include <iostream>
# Include <cstring>
# Include <fstream>
# Include <cstdlib>
# Include <memory>
# Include <algorithm>
# Define read freopen ("numtri. In", "r", stdin)
# Define write freopen ("numtri. Out", "W", stdout)
# Define RD ifstream cout ("numtri. Out ");
# Define wt ifstream CIN ("numtri. In ");
# Define max (A, B) A> B? A: B
Int data [1001] [1001];
Int main (){
Read;
Write;
Int I, j, N;
Scanf ("% d", & N );
For (I = 0; I <n; I ++)
For (j = 0; j <= I; j ++)
Scanf ("% d", & Data [I] [J]);
For (I = n-2; I> = 0; I --)
For (j = 0; j <= I; j ++)
Data [I] [J] + = max (data [I + 1] [J], data [I + 1] [J + 1]);
Printf ("% d \ n", data [0] [0]);
Return 0 ;}

 

2. It is obviously impossible to create a prime number table for this question, so we need to get a return number. Two methods:

(1) 10000000 judge in turn whether it is a return, of course, you can skip a lot in the middle, such as an even number, for example, the length except 11 is an even number.

(2) generate an odd number of input objects with an odd Length plus 11.

First, I wrote it in the first way. After writing it, I added a variety of optimizations and pruning. The Code is the same as the pus, but it still times out. The second method is much better. Although the code is ugly, the efficiency is quite high. Slowest 0.03

Code:

# Include <cstdlib>
# Include <cstring>
# Include <cstdio>
# Include <cmath>
# Include <algorithm>
# Include <string>
# Include <fstream>
# Define read freopen ("pprime. In", "r", stdin)
# Define write freopen ("pprime. Out", "W", stdout)
# Define RD ifstream cout ("pprime. Out ");
# Define wt ifstream CIN ("pprime. In ");
Int CNT, pa [1000001];
Void get_palin (){
Int I, J, K, L, G;
CNT = 0;
Pa [CNT ++] = 5;
Pa [CNT ++] = 7;
Pa [CNT ++] = 11;
For (I = 1; I <10; I + = 2)
For (j = 0; j <10; j ++)
Pa [CNT ++] = 100 * I + I + 10 * j;
For (I = 1; I <10; I + = 2)
For (j = 0; j <10; j ++)
For (k = 0; k <10; k ++)
Pa [CNT ++] = 10000 * I + I + 1000 * j + 10 * j + 100 * K;
For (I = 1; I <10; I + = 2)
For (j = 0; j <10; j ++)
For (k = 0; k <10; k ++)
For (L = 0; L <10; l ++)
Pa [CNT ++] = 1000000 * I + I + 100000 * j + 10 * j + 10000 * k + 100 * k + 1000 * l;
}
Int is_prim (int ){
For (INT I = 2; I <= SQRT (a); I ++)
If (a % I = 0) return 0;
Return 1;
}
Int main (){
Read;
Write;
Int I, n, m;
Scanf ("% d", & N, & M );
Get_palin ();
For (I = 0; I <CNT; I ++ ){
If (PA [I] <n | pa [I]> M |! Is_prim (PA [I]) continue;
Printf ("% d \ n", pa [I]);
}
Return 0;
}

 

3. Like the previous question, it is also directly generated and can be implemented recursively. I have written it as a.

Code:

# Include <cstdlib>
# Include <cstring>
# Include <cstdio>
# Include <cmath>
# Include <algorithm>
# Include <string>
# Include <fstream>
# Define read freopen ("sprime. In", "r", stdin)
# Define write freopen ("sprime. Out", "W", stdout)
# Define RD ifstream cout ("sprime. Out ");
# Define wt ifstream CIN ("sprime. In ");
Int ans [10] [1, 100001], CNT [11];
Int is_prim (int ){
For (INT I = 2; I <= SQRT (a); I ++)
If (a % I = 0) return 0;
Return 1;
}
Int main (){
Read;
Write;
Int N, I, J, K;
Scanf ("% d", & N );
Memset (CNT, 0, sizeof (CNT ));
CNT [1] = 4;
Ans [1] [0] = 2, ANS [1] [1] = 3, ANS [1] [2] = 5, ANS [1] [3] = 7;
For (I = 2; I <9; I ++)
For (k = 0; k <CNT [I-1]; k ++)
For (j = 1; j <10; j + = 2 ){
Int T = ans [I-1] [k] * 10 + J;
If (is_prim (t) ans [I] [CNT [I] ++] = T;
}
For (I = 0; I <CNT [N]; I ++)
Printf ("% d \ n", ANS [N] [I]);
Return 0 ;}

 

4. Queen eight, DFS + backtracking.

Code:

# Include <cstdio>
# Include <iostream>
# Include <cstring>
# Include <fstream>
# Include <cstdlib>
# Include <memory>
# Include <algorithm>
# Define read freopen ("checker. In", "r", stdin)
# Define write freopen ("checker. Out", "W", stdout)
# Define RD ifstream cout ("checker. Out ");
# Define wt ifstream CIN ("checker. In ");
Int N, C [15], D1 [30], D2 [30], ANS [15], CNT;
Void print (){
Int I;
For (I = 1; I <n; I ++)
Printf ("% d", ANS [I]);
Printf ("% d \ n", ANS [I]);
}
Void DFS (int r ){
If (r = n + 1 ){
CNT ++;
If (CNT <4)
Print ();
Return;
}
For (INT I = 1; I <= N; I ++ ){
If (C [I] | D1 [R + I] | D2 [R-I + N]) continue;
C [I] = 1;
D1 [R + I] = 1;
D2 [R-I + N] = 1;
Ans [R] = I;
DFS (R + 1 );
C [I] = 0;
D1 [R + I] = 0;
D2 [R-I + N] = 0;
}
}
Int main (){
Read;
Write;
Scanf ("% d", & N );
Memset (C, 0, sizeof (c ));
Memset (D1, 0, sizeof (D1 ));
Memset (D2, 0, sizeof (D2 ));
CNT = 0;
DFS (1 );
Printf ("% d \ n", CNT );
Return 0 ;}