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:
给定三个数组a,b,c,每个数组有若干个数(<=500个),再给定一个数s要你判断是否存在s=a[i]+b[j]+c[k],存在一组数就输出YES,一组都不存在就输出NO。
思路:
A+B = X - C
先把a数组和b数组中的数相加成一个ab[500×500]的数组,这样就相当于ab[i]+c[j]=s;再变形一下,ab[i]=c[j]+s,只要在ab数组里用二分查找看能否把c[j]+s找出来。
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #define N 505 5 6 __int64 ab[N * N]; 7 int num; 8 9 bool _search(__int64 x) 10 { 11 int i = 0, l = num - 1; 12 int mid; 13 while(i <= l){ 14 mid = (i + l) / 2; 15 if(ab[mid] == x) 16 return true; 17 else if(ab[mid] < x) 18 i = mid + 1; 19 else 20 l = mid - 1; 21 } 22 return false; 23 } 24 25 int main() 26 { 27 int n, m, l, kase = 0, s; 28 __int64 a[N], b[N], c[N], x; 29 while(scanf("%d%d%d", &n, &m, &l) == 3){ 30 kase++; num = 0; 31 for(int i = 0; i < n; i++) 32 scanf("%I64d", &a[i]); 33 for(int i = 0; i < m; i++) 34 scanf("%I64d", &b[i]); 35 for(int i = 0; i < l; i++) 36 scanf("%I64d", &c[i]); 37 38 for(int i = 0; i < n; i++) 39 for(int j = 0; j < m; j++) 40 ab[num++] = a[i] + b[j]; 41 std::sort(ab, ab+num); 42 std::sort(c, c+l); 43 44 scanf("%d", &s); 45 printf("Case %d: ", kase); 46 while(s--){ 47 scanf("%I64d", &x); 48 if(x < ab[0] + c[0] || x > ab[num-1] + c[l-1]) 49 printf("NO "); 50 else{ 51 int flag = 0; 52 for(int i = 0; i < l; i++){ 53 __int64 p = x - c[i]; 54 if(_search(p)){ 55 printf("YES "); 56 flag = 1; 57 break; 58 } 59 } 60 if(!flag) 61 printf("NO "); 62 } 63 } 64 } 65 return 0; 66 }