Can you find it?(二分 二分+STL set map)

Can you find it?

Time Limit : 10000/3000ms (Java/Other)   Memory Limit : 32768/10000K (Java/Other)

Total Submission(s) : 25   Accepted Submission(s) : 7

Problem 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

题意：分别有a,b,c三个数组，给出一个数，看这个数能不能再a,b,c,里面分别找出一个数求和得到

题解：先把两个数组加起来，另外一个数组用二分；

代码：

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;
 __int64 D[],A[],B[],C[],L,N,M;//D数组要开大；；；
 int erfen(__int64 *a,__int64 c,__int64 d){
     int l,r,mid;
     l=;r=L*N-;////////
     while(l<=r){
         mid=(l+r)/;
         if(a[mid]+c<d)l=mid+;
         else r=mid-;
         if(a[mid]+c==d)return ;//这点弄错了，交了几十遍。。。。。
     }
     return ;
 }
 int search(__int64 *a,__int64 *b,__int64 c){
     for(int i=;a[i];i++){
         if(erfen(b,a[i],c))return ;
     }
     return ;
 }
 void merge(__int64 *a,__int64 *b,__int64 *c){
     int t=;
     for(int i=;a[i];i++){
         for(int j=;b[j];j++){
             c[t++]=a[i]+b[j];
         }
     }
 }
 int main(){
     __int64 S,flot=,X;
     while(~scanf("%I64d%I64d%I64d",&L,&N,&M)){flot++;
         for(int i=;i<L;++i)scanf("%I64d",&A[i]);
         for(int i=;i<N;++i)scanf("%I64d",&B[i]);
         for(int i=;i<M;++i)scanf("%I64d",&C[i]);
         merge(A,B,D);
         sort(D,D+L*N);//这里也错了好多次，是相乘
             //for(int i=0;i<L+N;i++)printf("%d ",D[i]);
         scanf("%I64d",&S);
         printf("Case %I64d:\n",flot);
         while(S--){
             scanf("%I64d",&X);
             if(search(C,D,X))puts("YES");
             else puts("NO");
         }
     }
     return ;
 }

stl:

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;
 __int64 D[],A[],B[],C[],L,N,M;//D数组要开大；；；
 /*int erfen(__int64 *a,__int64 c,__int64 d){
     int l,r,mid;
     l=0;r=L*N-1;////////
     while(l<=r){
         mid=(l+r)/2;
         if(a[mid]+c<d)l=mid+1;
         else r=mid-1;
         if(a[mid]+c==d)return 1;//这点弄错了，交了几十遍。。。。。
     }
     return 0;
 }*/
 int search(__int64 *a,__int64 *b,__int64 c){
     for(int i=;a[i];i++){
         if(*lower_bound(b,b+L*N,c-a[i])==c-a[i])return ;/*lower_bound返回的是b数组中第一个大于等于c-a[i]元素的的地址
         upper(begin,end,index)找到大于某值的第一次出现；
         */
     }
     return ;
 }
 void merge(__int64 *a,__int64 *b,__int64 *c){
     int t=;
     for(int i=;a[i];i++){
         for(int j=;b[j];j++){
             c[t++]=a[i]+b[j];
         }
     }
 }
 int main(){
     __int64 S,flot=,X;
     while(~scanf("%I64d%I64d%I64d",&L,&N,&M)){flot++;
         for(int i=;i<L;++i)scanf("%I64d",&A[i]);
         for(int i=;i<N;++i)scanf("%I64d",&B[i]);
         for(int i=;i<M;++i)scanf("%I64d",&C[i]);
         merge(A,B,D);
         sort(D,D+L*N);//这里也错了好多次，是相乘
             //for(int i=0;i<L+N;i++)printf("%d ",D[i]);
         scanf("%I64d",&S);
         printf("Case %I64d:\n",flot);
         while(S--){
             scanf("%I64d",&X);
             if(search(C,D,X))puts("YES");
             else puts("NO");
         }
     }
     return ;
 }

map和set做全都MLE了，无奈了，别人的都对的；

set代码:

 #include<stdio.h>
 #include<set>
 using namespace std;
 const int MAXN=;
 int A[MAXN],B[MAXN],C[MAXN];
 int main(){
     int L,N,M,S,X,flot=,ok;
     while(~scanf("%d%d%d",&L,&N,&M)){set<int>num;
         for(int i=;i<L;++i)scanf("%d",&A[i]);
         for(int i=;i<N;++i)scanf("%d",&B[i]);
         for(int i=;i<M;++i)scanf("%d",&C[i]);
         for(int i=;i<L;i++)for(int j=;j<N;j++)num.insert(A[i]+B[j]);
         scanf("%d",&S);
         printf("Case %d:\n",flot++);
         while(S--){ok=;
             scanf("%d",&X);for(int i=;C[i];i++){
         if(num.find(X-C[i])!=num.end()){ok=;break;}
         }
             if(ok)puts("YES");
             else puts("NO");
         }
     }
     return ;
 } 

map代码：

 #include<stdio.h>
 #include<map>
 using namespace std;
 const int MAXN=;
 int A[MAXN],B[MAXN],C[MAXN];
 int main(){
     int L,N,M,S,X,flot=,ok;
     while(~scanf("%d%d%d",&L,&N,&M)){map<int,bool>num;
         for(int i=;i<L;++i)scanf("%d",&A[i]);
         for(int i=;i<N;++i)scanf("%d",&B[i]);
         for(int i=;i<M;++i)scanf("%d",&C[i]);
         for(int i=;i<L;i++)for(int j=;j<N;j++)num[A[i]+B[j]]=;
         scanf("%d",&S);
         printf("Case %d:\n",flot++);
         while(S--){ok=;
             scanf("%d",&X);for(int i=;C[i];i++){
         if(num.count(X-C[i])){ok=;break;}
         }
             if(ok)puts("YES");
             else puts("NO");
         }
     }
     return ;
 }