原题链接: 点我转移 https://vjudge.net/problem/18481/origin
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.
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;
- }
来源: http://www.bubuko.com/infodetail-2967882.html