HDU2116 Has the sum exceeded

Has the sum exceeded

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3218    Accepted Submission(s): 683

Problem Description

As we all know, in the computer science, an integer A is in the range of 32-signed integer, which means the integer A is between -2^31 and (2^31)-1 (inclusive), and A is a 64-signed integer, which means A is between -2^63 and (2^63)-1(inclusive). Now we give the K-signed range, and two K-signed integers A and B, you should check whether the sum of A and B is beyond the range of K-signed integer or not.

 

Input

There will be many cases to calculate. In each case, there comes the integer K (2<=K<=64) first in a single line. Then following the line, there is another single line which has two K-signed integers A and B.

 

Output

For each case, you should estimate whether the sum is beyond the range. If exceeded, print “Yes”, otherwise “WaHaHa”.

 

Sample Input

  
  

   
   32
100 100
  
  

 

Sample Output

  
  

   
   WaHaHa
  
  

 主要思路是把数转化为二进制判断

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
__int64 a[70];
int main()
{

    __int64 c,d;
    int num,k,i,j,flag;
    while(scanf("%d",&k)!=EOF)
    {
        flag=0;
        scanf("%I64d%I64d",&c,&d);
        num=0;
        memset(a,0,sizeof(a));
        if((c<0&&d>0)||(c>0&&d<0))
        {
            printf("WaHaHa\n");
            continue;
        }

         if(c<0&&d<0)
        {
            c=-c;
            d=-d;
            flag=1;
        }

        while(c!=0)
        {
           a[num++]=c%2;
           c=c/2;
        }

        num=0;
        while(d!=0)
        {
            a[num]=a[num]+d%2;
            num++;
            d=d/2;
        }

        for(i=0;i<68;i++)
        {
            a[i+1]=a[i+1]+a[i]/2;
            a[i]=a[i]%2;
        }
        for(i=68; i>=0; i--)
        if(a[i]!=0)
        break;
        
        
        if(i==k-1&&flag==1)
        {
            for(j=i-1;j>=0;j--)
            {
                if(a[j]!=0)
                break;
            }
        }
        if(i<k-1)
        printf("WaHaHa\n");
        else

        if(i==k-1&&flag==1&&j<0)
        {
             printf("WaHaHa\n");
        }
        else
        printf("Yes\n");
        }
    return 0;
}