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[1000010],A[510],B[510],C[510],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=0;a[i];i++){
        if(erfen(b,a[i],c))return 1;
    }
    return 0;
}
void merge(__int64 *a,__int64 *b,__int64 *c){
    int t=0;
    for(int i=0;a[i];i++){
        for(int j=0;b[j];j++){
            c[t++]=a[i]+b[j];
        }
    }
}
int main(){
    __int64 S,flot=0,X;
    while(~scanf("%I64d%I64d%I64d",&L,&N,&M)){flot++;
        for(int i=0;i<L;++i)scanf("%I64d",&A[i]);
        for(int i=0;i<N;++i)scanf("%I64d",&B[i]);
        for(int i=0;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:
",flot);
        while(S--){
            scanf("%I64d",&X);
            if(search(C,D,X))puts("YES");
            else puts("NO");
        }
    }
    return 0;
}

 stl:

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
__int64 D[1000010],A[510],B[510],C[510],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=0;a[i];i++){
        if(*lower_bound(b,b+L*N,c-a[i])==c-a[i])return 1;/*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 0;
}
void merge(__int64 *a,__int64 *b,__int64 *c){
    int t=0;
    for(int i=0;a[i];i++){
        for(int j=0;b[j];j++){
            c[t++]=a[i]+b[j];
        }
    }
}
int main(){
    __int64 S,flot=0,X;
    while(~scanf("%I64d%I64d%I64d",&L,&N,&M)){flot++;
        for(int i=0;i<L;++i)scanf("%I64d",&A[i]);
        for(int i=0;i<N;++i)scanf("%I64d",&B[i]);
        for(int i=0;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:
",flot);
        while(S--){
            scanf("%I64d",&X);
            if(search(C,D,X))puts("YES");
            else puts("NO");
        }
    }
    return 0;
}

 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=505;
int A[MAXN],B[MAXN],C[MAXN];
int main(){
    int L,N,M,S,X,flot=0,ok;
    while(~scanf("%d%d%d",&L,&N,&M)){set<int>num;
        for(int i=0;i<L;++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]);
        for(int i=0;i<L;i++)for(int j=0;j<N;j++)num.insert(A[i]+B[j]);
        scanf("%d",&S);
        printf("Case %d:
",flot++);
        while(S--){ok=0;
            scanf("%d",&X);for(int i=0;C[i];i++){
        if(num.find(X-C[i])!=num.end()){ok=1;break;}
        }
            if(ok)puts("YES");
            else puts("NO");
        }
    }
    return 0;
} 

map代码：

#include<stdio.h>
#include<map>
using namespace std;
const int MAXN=505;
int A[MAXN],B[MAXN],C[MAXN];
int main(){
    int L,N,M,S,X,flot=0,ok;
    while(~scanf("%d%d%d",&L,&N,&M)){map<int,bool>num;
        for(int i=0;i<L;++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]);
        for(int i=0;i<L;i++)for(int j=0;j<N;j++)num[A[i]+B[j]]=1;
        scanf("%d",&S);
        printf("Case %d:
",flot++);
        while(S--){ok=0;
            scanf("%d",&X);for(int i=0;C[i];i++){
        if(num.count(X-C[i])){ok=1;break;}
        }
            if(ok)puts("YES");
            else puts("NO");
        }
    }
    return 0;
}