HDOJ--2141 Can you find it?（包含题意）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17547    Accepted Submission(s): 4437

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO
题意：首先输入l,m,n,然后第一行输入l个整数，第二行输入m个整数，第三行输入n个整数。最终就是看你所给的数据中有没有满足第一行的一个数+第二行的一个数+第三行的一个数相等的结果，有的话就是"YES",没有就是NO.
思路：
主要就是利用二分法来进行时间的缩短免得超时。祥见代码。
错误：总之犯了两个错误，导致一道题做了一下午。1、就是没有把top初始化。2、输出个事不对，没有加“：“超坑人，一个冒号导致了无数次的wa.
ac代码：
#include<stdio.h>#include<algorithm>using namespace std;int a[550],b[550],c[550],s[550*550];int part(int l,int r,int path){    while(l<=r){            int mid=(l+r)/2;        if(s[mid]<path)            l=mid+1;        else if(s[mid]>path)            r=mid-1;        else            return 1;    }    return 0;}int main(){    int l,m,n,i,j,k,top,count=0,flag=0;    while(scanf("%d%d%d",&l,&n,&m)!=EOF){            top=-1;        for(i=0;i<l;i++)            scanf("%d",&a[i]);        for(j=0;j<m;j++)            scanf("%d",&b[j]);        for(k=0;k<n;k++)            scanf("%d",&c[k]);        for(i=0;i<l;i++)            for(j=0;j<m;j++)                s[++top]=a[i]+b[j];        sort(s,s+top+1);        int t;        scanf("%d",&t);        printf("Case %d:\n",++count);        while(t--){            int x;            flag=0;            scanf("%d",&x);            for(i=0;i<n;i++){                int y=x-c[i];                if(part(0,top,y)){                    flag=1;                    break;                }                    }            if(flag)                printf("YES\n");            else                printf("NO\n");        }    }}