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
AC代码：
#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>//排序的头文件
using namespace std;
int a[501],b[501],c[501],d[501*501];//合并数组的范围应为之前数组的平方
int main()
{
int l,n,m,i,j,k,p,y,s,x;
s=0;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
for(i=0;i<l;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
for(i=0;i<m;i++)
scanf("%d",&c[i]);
sort(c,c+m);//将数组进行排序
int q=0;
for(j=0;j<n;j++)
for(k=0;k<m;k++)
d[q++]=a[j]+b[k];//将前两个数组求和，合并成为一个数组
sort(d,d+q);
scanf("%d",&p);
printf("Case %d:\n",++s);//此处少写了换行符
for(i=0;i<p;i++)
{
scanf("%d",&x);
int flag=0;//标记
for(k=0;k<m;k++)
{
y=x-c[k];
int le,r,mid;
le=0;//左端点
r=q-1;//右端点,刚写错了，将区间的范围缩小了
while(le<=r)//二分法核心代码
{
mid=(le+r)/2;//中区间
if(y==d[mid])
{
flag=1;
break;//终止判断
}
if (d[mid]<y)
le=mid+1;
else r=mid-1;
}
if(flag==1) break;//如果有一个合法的y就结束循环
}
if(flag)
printf("YES\n");
else printf("NO\n");
}
}
return 0;
}
分析：就是给你ABC三个数，这三个数每一个都有很多的取值，现在给你一个x让你判断任意的ABC相加是否等于X（有多组值），若等于就YES，否则就NO，一开始我想用四个for循环来写但是后来想了一下觉得太繁琐而且又容易超时，于是找到了二分法来写，简单易懂。
具体步骤：将前两个数组相加后的值赋给一个新的数组，然后用x的值减去另外一个数组的值，判断两者是否相等。