Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 7898 Accepted Submission(s): 2052
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
Author
wangye
Source
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威士忌
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <cstdlib> 6 #include <algorithm> 7 #include <vector> 8 #include <stack> 9 #include <queue> 10 #include <cassert> 11 #include <set> 12 #include <sstream> 13 #include <map> 14 using namespace std ; 15 #ifdef DeBUG 16 #define bug assert 17 #else 18 #define bug // 19 #endif 20 #define zero {0} 21 const int MAXN=1000; 22 int L,M,N; 23 int a[MAXN],b[MAXN],c[MAXN]; 24 int ab[300000]; 25 int S,n; 26 int cmp(const void*a,const void*b) 27 { 28 return *(int *)a-*(int *)b;//升序 29 } 30 int find(int min,int max,int x) 31 { 32 int mid; 33 while(min<=max) 34 { 35 mid=(min+max)/2; 36 if(ab[mid]==x) 37 return 1; 38 if(ab[mid]>x) 39 max=mid-1; 40 if(ab[mid]<x) 41 min=mid+1; 42 } 43 return 0; 44 } 45 int main() 46 { 47 #ifdef DeBUG 48 freopen("C:\Users\Sky\Desktop\1.in","r",stdin); 49 #endif 50 51 int i,j,k; 52 int cas=1; 53 int num; 54 while(scanf("%d%d%d",&L,&N,&M)!=EOF) 55 { 56 num=0; 57 for(i=0;i<L;i++) 58 scanf("%d",&a[i]); 59 for(i=0;i<N;i++) 60 scanf("%d",&b[i]); 61 for(i=0;i<M;i++) 62 scanf("%d",&c[i]); 63 scanf("%d",&S); 64 for(i=0;i<L;i++)//将两个数组合并成一个数组，使其为两个数组中各个不同元素的和 65 for(j=0;j<N;j++) 66 ab[num++]=a[i]+b[j]; 67 // qsort(c,M,sizeof(c[0]),cmp);//可排可不排 68 qsort(ab,num,sizeof(ab[0]),cmp);//二分查找的前提是有序的数组 69 printf("Case %d: ",cas++); 70 while(S--&&scanf("%d",&n)) 71 { 72 bool _flag=false; 73 for(i=0;i<M;i++) 74 { 75 if(find(0,num-1,n-c[i])) 76 { 77 _flag=true; 78 break; 79 } 80 } 81 if(_flag) 82 printf("YES "); 83 else 84 printf("NO "); 85 } 86 } 87 return 0; 88 }