HDU 2141 Can you find it?

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 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output
Case 1:
NO
YES
NO

题意为，给出多个a,b,c的值和x的值，判断是否有一组a,b,c的和等于x

大致思路为，三重循环超时，用空间换时间，将a+b+c=x换做(a+b)+c=x，也就是a+b=x-c。先求出每一种a+b的和的可能性存入数组，并排序。二分的上下界为数组下标，0到总共可能性的个数。二分来查找是否存在a+b等于x-c的值。对于每个x，遍历每一个c值，看是否存在一个c，使得a+b=x-c。如果没有，则输出no，否则输出yes。

遇到的问题，在最后判断输出no时。应当在遍历每一个c都不符合后，判断应当为if(i>=m)。

代码如下：

#include<stdio.h>#include<iostream>#include<cstring>#include<string.h>#include<cmath>#include<algorithm>using namespace std;int sum[250009],a[509],b[509],c[509],l,n,m,s,x;bool bi_search(int rst){    int le=0,ri=l*n-1;    while(le<=ri){        int mid=(le+ri)/2;        if(sum[mid]==rst)            return true;        else if(sum[mid]>rst)            ri=mid-1;        else            le=mid+1;    }    return false;}int main(){    int cnt =1;    while(~scanf("%d%d%d",&l,&n,&m)){        int cn=0; //cn记录sum数组里的个数，cnt记录case数        for(int i=0;i<l;i++)            scanf("%d",&a[i]);        for(int i=0;i<n;i++)            scanf("%d",&b[i]);        for(int i=0;i<m;i++)            scanf("%d",&c[i]);        for(int i=0;i<l;i++)            for(int j=0;j<n;j++)                sum[cn++]=a[i]+b[j];        sort(sum,sum+cn);        scanf("%d",&s);        printf("Case %d:\n",cnt++);        while(s--){            scanf("%d",&x);            int i;            for(i=0;i<m;i++){                if(bi_search(x-c[i])){                    puts("YES");                    break;                }            }            if(i>=m)                puts("NO");        }    }    return 0;}