二分贪心专题C

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

题目大意：给你三个数组A，B，C。问能不能再A，B，C数组中各找一个数使得他们的和为X里。

从数据范围来看A，B，C数组都不超过500个，但是暴力的话复杂度为n^3 空间与时间复杂度难以接受。

换个思路，将A，B数组合并为数组D，那么问题就变成了Di+Ci=X

即X-Di=Ci

枚举D数组中的元素并在C数组中进行二分查找。

依然是思路的问题。我们看到500^3难以接受，但是500*500很好实现，自己的榆木脑袋还是有待雕琢呀。。。

源代码：

#include<iostream>#include <algorithm>using namespace std;int main(){int l,n,m,total=0;while (cin>>l>>n>>m){total++;int tot=0,s;int a[505];int b[505];int c[505];int d[260000];for (int i=1;i<=l;i++) cin>>a[i];for (int i=1;i<=n;i++) cin>>b[i];for (int i=1;i<=m;i++) cin>>c[i];cin>>s;for (int i=1;i<=l;i++)for (int j=1;j<=n;j++) {tot++;d[tot]=a[i]+b[j];}sort(d+1,d+tot+1);cout<<"Case "<<total<<":"<<endl;for (int i=1;i<=s;i++){bool flag=false;int k;cin>>k;for (int j=1;j<=m;j++){int l=1,r=tot,mid;int ans=k-c[j];while (l<=r){mid=(l+r)/2;if (d[mid]==ans) {flag=true;break;}if (d[mid]<ans) l=mid+1;if (d[mid]>ans) r=mid-1;}if (flag) break;}if (flag) cout<<"YES"<<endl;else cout<<"NO"<<endl;}}return 0;}