Search deep pruning

Deep search has been mentioned, for pruning is to carry out the necessary steps if no special pruning sentence, with good complexity can greatly optimize

According to two examples here in terms of it

1440: division [1] number of examples

For details, see the code

#include <cstdio> 
#include <CString> 
#include <algorithm>
 int n-, K;
 int F [ 210 ] [ . 7 ];
 int main () 
{ 
    Scanf ( " % D% D " , & n-, & K); 
    Memset (F, 0 , the sizeof (F)); // F is initialized array 0, 
    F [ 0 ] [ 0 ] = 1 ; // but the first to 1, in order to prevent Calais are added to 0 
    for ( int I = . 1 ; I <= n-; I ++) // sum cycle 
        for (int j = I; j <= n-; j ++) // is the sum of the cycle, as has already been cycled i, j may thus directly start I
         // two cycles later to pave the way for the state of the processing 
            for ( int X = . 1 ; X <= K; X ++) // this is the number of cycles of the program 
            { 
                F [J] [X] = F [J] [X] + F [JI] [X- . 1 ]; // F [J] [x] is the maximum value of the total number of programs
                 @ JI is to prevent repeated, for example: 1 1 1 1 1,5 5,1 5 this happens
                 // is a maximum value of x-1, in order to inherit a 
            } 
    the printf ( " % D \ n- " , F [n-] [K]); // output maximum 
    return  0 ; 
}

1441: example [2] birthday cake

For details, see the code

include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
using namespace std;
const int maxn=0x3f3f3f3f;
const int minn=-0x3f3f3f3f;
int n,m,ans,a[10086],b[10086];
inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x *F; 
} 
/ * 
Volume V = πR2H 
side area A = 2πRH 
bottom area πR2 = A 
* / 
void DFS ( int V / * Used by volume * / , int S / * surface area * / , int P / * remaining layers Note residual * / , int R & lt / * radius * / , int H / * high * / ) {
     IF (P == 0 ) { // If no remaining layers has been found the search, then no! 
        IF (V n-S && == < ANS) 
            ANS =S;
         return ; 
    } 
    IF (V + B [p- . 1 ]> n)
         return ; // ! add if the volume is greater than the maximum volume of this layer is further found that the n ass 
    IF (A + S [p- . 1 ]> ANS)
         return ;
     IF ( 2 * (NV) / R & lt S +> = ANS)
         return ; // current surface area of the remaining lateral area +> further found that this optimum value ass! 
    for ( int I = R- . 1 ; I> = P; i--) { // radius 
        IF (P == m) 
             S = I * I;
         int pyyyyyy = min ((NVB [p-1])/(i*i),h-1);
        
        for(int j=pyyyyyy; j>=p; j--) //高 
            dfs(v+i*i*j,s+2*i*j,p-1,i,j);
    }
}
int main() {
    n=read();
    m=read();
    cin>>n>>m;
    ans=maxn;
    a[0]=0;
    b[0]=0;
    for(int i=1; i<21; i++) { 
        A [i] = A [I- . 1 ] + 2 * i * i; // maximum surface area layer i 
        B [i] = B [I- . 1 ] + i * i * i; // i layer volume = πR2H maximum volume V 
    } 
    DFS ( 0 , 0 , m, n-+ . 1 , n-+ . 1 ); // search
     @ 5 mean the amount of fancy 
    IF (ANS == MAXN) 
        COUT << 0 ;
     the else 
        COUT < < ANS;
     return  0 ; 
}