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 3 1 2 3 1 2 3 1 2 3 3 1 4 10
Sample Output
Case 1: NO YES NO
题目大意及分析
给定三个数组a，b，c，每个数组有若干个数（<=500个），再给定一个数s要你判断是否存在s=a[i]+b[j]+c[k]，存在一组数就输出YES，一组都不存在就输出NO。
因为只有三个500大小的数组，刚开始我直接写了三个for循环，超时，再看看题目就更清楚了，每组测试数据中给的s是<=1000个，这就很容易超时，而且，给的数可以是负数，这样我sort全部排序，也无济于事。
无奈之下直接百度，才发现已经AC的人都先把a数组和b数组中的数相加成一个ab[500×500]的数组，这样就相当于ab[i]+c[j]=s;再变形一下，ab[i]=c[j]+s;这样我就知道接下来要干什么了，只要在ab数组里用以前学的二分查找看能否把c[j]+s找出来，貌似以前也有过这类要把数据关系进行转换的题。
代码如下：
#include<cstdio>
#include<cstdlib>
int a[501],b[501],c[501],d[260000];int k,n,m;
int cmp(const void*a,const void*b)
{
return *(int*)a-*(int*)b;
}
int main(void)
{
int t,s,mm=1;
while(scanf("%d%d%d",&k,&n,&m)!=EOF)
{
for(int i=0;i<k;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]);
int cnt=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
d[cnt++]=b[i]+c[j];
qsort(a,k,sizeof(a[0]),cmp);
qsort(d,cnt,sizeof(d[0]),cmp);
scanf("%d",&t);
printf("Case %d:\n",mm++);
while(t--)
{
int flag=0;
scanf("%d",&s);
for(int i=0;i<k;i++)
{
int tl=0,tr=cnt-1;
int tt=s-a[i];
while(tl<tr)
{
int mid=(tl+tr)/2;
if(d[tl]==tt || d[tr]==tt || d[mid]==tt){flag=1;break;}
else if(d[mid]<tt)tl=mid+1;
else tr=mid-1;
}
if(flag==1) break;
}
if(flag==1) puts("YES");
else if(flag==0) puts("NO");
}
}
return 0;
}