Hdu 2141 binary

Hdu 2141 binary

The meaning is to give you three arrays with n m k numbers, respectively. Can we take a number from the three groups to be equal to s?

At the beginning, I didn't pay attention to the fact that the number can be negative wa many times.

Don't try brute force. It must have timed out.

Merge the two arrays into a list with a smaller number of brute force enumerations)

#include
 
  #include
  
   #include
   
    using namespace std
    ;int main(){    int num1
    [510
    ],num2
    [510
    ],num3
    [510
    ];    int mark
    [300000
    ];    int i
    ,j
    ,k
    ,p
    ,d
    =1
    ;    int n
    ,m
    ;    while(~scanf
    ("%d%d%d"
    ,&n
    ,&m
    ,&k
    ))    {        for(i
    =1
    ;i
    <=n
    ;i
    ++)        scanf
    ("%d"
    ,&num1
    [i
    ]);        for(i
    =1
    ;i
    <=m
    ;i
    ++)        scanf
    ("%d"
    ,&num2
    [i
    ]);        p
    =0
    ;        for(i
    =1
    ;i
    <=n
    ;i
    ++)        for(j
    =1
    ;j
    <=m
    ;j
    ++)        mark
    [++p
    ]=num1
    [i
    ]+num2
    [j
    ];        sort
    (mark
    +1
    ,mark
    +1
    +p
    );        for(i
    =1
    ;i
    <=k
    ;i
    ++)        scanf
    ("%d"
    ,&num3
    [i
    ]);        sort
    (num3
    +1
    ,num3
    +1
    +k
    );        int s
    ;        scanf
    ("%d"
    ,&s
    );        printf
    ("Case %d:\n"
    ,d
    ++);        while(s
    --)        {            int sum
    ;            scanf
    ("%d"
    ,&sum
    );            int flash
    =0
    ;            int left
    ,right
    ,mid
    ;            for(i
    =1
    ;i
    <=k
    ;i
    ++)            {                left
    =1
    ;                right
    =p
    ;                while(left
    <=right
    )                {                    mid
    =(left
    +right
    )/2
    ;                    if(mark
    [mid
    ]>sum
    -num3
    [i
    ]) right
    =mid
    -1
    ;                    else if(mark
    [mid
    ]<sum
    -num3
    [i
    ]) left
    =mid
    +1
    ;                    else                     {                        flash
    =1
    ;                        break;                    }                }                if(flash
    ) break;            }            if(flash
    ) printf
    ("YES\n"
    );            else printf
    ("NO\n"
    );        }    }    return 0
    ;}