题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141
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 10Sample Output
Case 1: NO YES NO
这道题的题意是有三个数组,从每个数组中找到一个数据加起来等于下面的数字。
如果暴力来做思路很简单,三个for循环就可以了,但是这道题的时间范围有要求,暴力绝对超时,用二分就避免了这种问题。
代码如下:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
ll a[505],b[505],c[505],d[250005];
int l,n,m,s,k=0,ca=1;
int pan(ll x)//核心代码,二分遍历
{
int t=0,y=k-1;
while(t<=y)
{
int mi=(t+y)/2;
if(d[mi]==x)
return 1;
else if(d[mi]>x)
y=mi-1;
else
t=mi+1;
}
return 0;
}
int main()
{
while(~scanf("%d%d%d",&l,&n,&m))
{
for(int i=0; i<l; i++)
scanf("%lld",&a[i]);
for(int i=0; i<n; i++)
scanf("%lld",&b[i]);
for(int i=0; i<m; i++)
scanf("%lld",&c[i]);
k=0;
for(int i=0; i<l; i++)//把前两个数组相加
{
for(int j=0; j<n; j++)
d[k++]=a[i]+b[j];
}
sort(d,d+k);
ll sa;
scanf("%d",&s);
printf("Case %d:\n",ca++);
while(s--)
{
int flag=0;
scanf("%lld",&sa);
for(int i=0; i<m; i++)
{
if(pan(sa-c[i]))
{
flag=1;
break;
}
}
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
}
}
这道题就是基础的二分省时,我自己也没想到,还是看了学长的代码才发现,自己水平确实下降了。