原题链接：点我转移
InputThere 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.
OutputFor 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
#include<bits/stdc++.h> #define INF 0x3f3f3f3f #define PI 3.1415926 using namespace std; typedef long long ll; const int maxn=500+7; const ll mod=1e9+7; int main() { int a[maxn],b[maxn],c[maxn],sum[maxn*maxn]; int l,n,m; int t=1; while(cin>>l>>n>>m) { for(int i=0; i<l; i++) cin>>a[i]; for(int i=0; i<n; i++) cin>>b[i]; for(int i=0; i<m; i++) cin>>c[i]; for(int i=0; i<l; i++) for(int j=0; j<n; j++) sum[i*l+j]=a[i]+b[j]; sort(sum,sum+l*n); cout<<"Case "<<t++<<":"<<endl; int s; cin>>s; while(s--) { int flag=0; int r; cin>>r; for(int i=0; i<m; i++) if(binary_search(sum,sum+l*n,r-c[i])) { flag=1; break; } if(flag) cout<<"YES"<<endl; else cout<<"NO"<<endl; } } return 0; }
原文：https://www.cnblogs.com/tp25959/p/10434605.html