【HDU】-2141-Can you find it?（二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 23795    Accepted Submission(s): 6032

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

3 3 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

先把a，b数组合并，存入数组d。排序，再用一个n=unique(a,a+n)-a的函数去除掉d里相同的数,因为不去除的话数会有点多,
(函数实际操作：调试表明不同所有数在前,相同在后),n的值改变，变为前面不同的数的个数。
再把c数组里的数过一遍，对d数组二分，可以用已有函数（lower_bound和upper_bound），找到d中与c[i]的和是所求的数，标记，退出。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std; int main(){int x,y,z,num=1,ant;int a[510],b[510],c[510],d[250010];while(~scanf("%d %d %d",&x,&y,&z)){ant=0;for(int i=1;i<=x;i++)scanf("%d",&a[i]);for(int i=1;i<=y;i++){scanf("%d",&b[i]);for(int j=1;j<=x;j++){d[ant++]=a[j]+b[i];//将a,b数组合并，记为d }}sort(d,d+ant);ant=unique(d,d+ant)-d; //去除d中相同数，保留不同的，此时d已经重新排序(调试表明不同所有数在前,相同在后),ant的值为前面不同的个数 for(int i=1;i<=z;i++)scanf("%d",&c[i]);printf("Case %d:\n",num++);int t;scanf("%d",&t);while(t--){int m,flag=0;scanf("%d",&m);for(int i=1;i<=z;i++){int l,r,mid;l=d[0],r=d[ant-1];int pos=lower_bound(d,d+ant,m-c[i])-d;  //d是递增的数组，找到和m-c[i]相同的值减去首地址返回下标 if(d[pos]+c[i]==m)//d数组必须从0开始，因为用已知函数如果找不到符合的返回b[0]，而b[0]=0加c[i]可能与m相同 {flag=1;break;}}if(flag)printf("YES\n");elseprintf("NO\n");}}return 0;}