dinner party dp 暴力

dp用bitset處理

讀取時注意判斷

#include <bits/stdc++.h>
using namespace std ;
typedef long long ll ;

    /*bitset<32> bitvec2(0xffff); // bits 0 ... 15 are set to 1; 16 ... 31 are 0
    cout << "bitvec2: " << bitvec2 << endl;
    bitset<32> a(3) ;
    cout << a << endl ;*/
    ///bitset is very useful

const int N = 1010 , M = 2010 ;
int T,n,m,i,j,x,y,cnt;
bitset<M>f[N];
void solve(){
    scanf("%d %d" , &n , &m) ;
    if(m % 2){
        puts("No") ;
        return ;
    }
    m /= 2 ;
    if( !f[n][m] ){
        puts("No") ;
        return ;
    }
    puts("Yes") ;
    int cnt = 0 ;
    //static
    int q[100000][2] ;
    while( n ){
        int X = 0 , Y = 0 ;
        for(x = 1 ; x <= n ; x ++ ){
            for(y = 1 ; y * x <= n ; y ++ ) if(x + y <= m ) if(f[n - x * y][m - x - y ]){
                X = x , Y = y ;
                break ;
            }if(X) break ;
        }
        q[ ++ cnt][0] = X ;
        q[cnt][1] = Y ;
        n -= X * Y ;
        m -= X + Y ;
    }
    printf("%d\n" , cnt) ;
    for(int i = 1 ; i <= cnt ; i ++ ) printf("%d %d\n" , q[i][0] , q[i][1]) ;
}
int main(){
    n = 1000 ;
    m = 2000 ;
    f[0][0] = 1 ;
    for(i = 0 ; i <= n ; i ++ )
        for(x = 1 ; x + i <= n ; x ++ )
            for(y = 1 ; x * y + i <= n ; y ++ )
                f[i + x * y] |= f[i] << (x + y) ;
    //printf("ashgfkla") ;
    //for(int i = 0 ; i < 1 ; i ++ )
      //  cout << f[i] << endl ;
    scanf("%d" ,&T) ;
    while( T -- ) solve() ;
}