hdu2241（二分+合并数组）

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题意：判断是否存在a+b+c = x，a，b，c分别属于集合A,B,C

如果用暴力会超时，所以这里用到了数组合并，将b,c数组合并成d，d数组存的是b，c数组元素的和，然后对d数组进行二分就可以了

代码如下（附注释）：

#include<iostream>#include<algorithm>#include<cstring>#include<stack>#include<queue>#include<set>#include<map>#include<stdio.h>#include<stdlib.h>#include<ctype.h>#include<time.h>#include<math.h>#define inf 0x7fffffff#define eps 1e-9#define N 505#define pi acos(-1.0)#define P system("pause")using namespace std;int a[N],b[N],c[N],d[N*N];int main(){//freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);    int l,m,n,z = 1;    int i,j,k;    while(scanf("%d%d%d",&l,&m,&n) != EOF)    {          for(i = 0;i < l; i++)                scanf("%d",&a[i]);          for(i = 0; i < m; i++)                scanf("%d",&b[i]);          for(i = 0; i < n; i++)                scanf("%d",&c[i]);           k = 0;           for(i = 0; i < m; i++)//将b,c两个数组和并在d数组内              for(j = 0; j < n; j++)                 d[k++] = b[i] + c[j];          sort(d,d+k);          j = 1;          for(i = 1; i < k; i++)//对d数组就行优化，去掉相同的数据                 if(d[i] != d[i-1])                   d[j++] = d[i];         // for(i = 0; i < j; i++)           //  cout<<d[i]<<" ";c                                                     int t, flag;         scanf("%d",&t);         printf("Case %d:\n",z++);         while(t--)         {             flag = 0;                   scanf("%d",&k);             for(i = 0; i < l; i++)             {                   int left = 0,right = j-1;                   int k1 = k - a[i];                  // cout<<k<<endl;                   while(left <= right)//二分查找                    {                        int mid = (left + right)/2;                        if(d[mid] == k1)                        {                               printf("YES\n");                               i = l;                               flag = 1;                               break;                                         }                         if(d[mid] < k1) left = mid + 1;                        if(d[mid] > k1) right = mid - 1;                   }              }                 if(!flag) printf("NO\n");                      }              }                                           // P;                                   return 0;    }

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