Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 35782 Accepted Submission(s): 8831
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
思路：
此处可以用暴力求解法以及二分查找法（会节省更多时间）。而二分查找是有对应的STL的。此处是可以用binary_search,是在有序数列中确定给定的元素是否存在。
#include <iostream> #include <algorithm> using namespace std; int a[501],b[501],c[501],sum[250010],s[1001]; int main() { int l,n,m,i,j,s1,T=0; while(scanf("%d%d%d",&l,&n,&m)!=EOF) { T++; for(i=0;i<l;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) scanf("%d",&b[i]); for(i=0;i<m;i++) scanf("%d",&c[i]); for(i=0;i<l;i++) { for(j=0;j<n;j++) sum[n*i+j]=a[i]+b[j]; } sort(sum,sum+l*n); scanf("%d",&s1); for(i=0;i<s1;i++) scanf("%d",&s[i]); printf("Case %d:\n",T); for(i=0;i<s1;i++) { for(j=0;j<m;) { if(binary_search(sum,sum+l*n,s[i]-c[j])) break; j++; } if(j!=m) printf("YES\n"); else printf("NO\n"); } } return 0; }
转载于:https://www.cnblogs.com/Weixu-Liu/p/9173673.html
相关资源：JAVA上百实例源码以及开源项目