题目链接:clicl here~~
【题目大意】:
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.
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
if(sort[mid]<ans) left=mid+1;
else if(sort[mid]>ans) right=mid-1;
else {flag=true;break;}
代码:
#include <bits/stdc++.h>
using namespace std;
const int N=505;
int A[N],B[N],C[N],D[N*N];
bool get(int sort[],int k,int ans)//合并后的数组/数组元素个数/和减去第三个数组元素剩下的值
{
bool flag=false;
int left=0,right=k,mid;
while(left<=right)
{
mid=(left+right)>>1;
if(sort[mid]<ans) left=mid+1;
else if(sort[mid]>ans) right=mid-1;
else {flag=true;break;}
}
return flag;
}
int main()
{
//freopen("1.txt","r",stdin);
int l,n,z,m,S,tot=1;
bool ok;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
for(int i=0;i<l;i++) scanf("%d",&A[i]);
for(int i=0;i<n;i++) scanf("%d",&B[i]);
for(int i=0;i<m;i++) scanf("%d",&C[i]);
int k=0;
for(int i=0;i<l;i++)
for(int j=0;j<n;j++) D[k++]=A[i]+B[j];
sort(D,D+k);
scanf("%d",&S);
printf("Case %d:\n",tot++);
while(S--)
{
ok=false;
scanf("%d",&z);
for(int i=0;i<m;i++){
if(get(D,k,z-C[i])) {ok=true;break;}
}
if(ok) puts("YES");
else puts("NO");
}
}
return 0;
}