HDU 2141 Can you find it? <二分>
Can you find it?
Time Limit:3000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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
Sample Input
Sample Output
#include<cstdio>
#include<cctype>
#include<iostream>
#include<stack>
#include<map>
#include<cstring>
#include<string>
#include<sstream>
#include<queue>
#include<set>
using namespace std ;
int a[505],b[505],c[505];
set<int>sum;
int main() {
int l,n,m;
int cou=1;
while(~scanf("%d%d%d",&l,&n,&m)){
sum.clear();
for(int i=0;i<l;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int j=0;j<m;j++)
scanf("%d",&c[j]);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++){
sum.insert(b[i]+c[j]);
}
int t;
scanf("%d",&t);
printf("Case %d:\n",cou++);
while(t--){
int tp;
scanf("%d",&tp);
int flag=0;
set<int>::iterator f;
for(int i=0;i<l;i++){
f=sum.find(tp-a[i]);//利用find函數來搜索值，其實set集合類的find函數內部就是用二分來實現的，所以想要快速AC的同學，可以使用這種方法.
if(f!=sum.end()){
flag=1;
break;
}
}
if(!flag)
printf("NO\n");
else
printf("YES\n");
}
}
return 0 ;
}
#include<cstdio>
#include<cctype>
#include<iostream>
#include<stack>
#include<map>
#include<cstring>
#include<string>
#include<sstream>
#include<queue>
#include<set>
#include<algorithm>
using namespace std ;
int a[505],b[505],c[505];
int sum[250005];
int k=0,flag=0;
void bs(int tp,int i)
{
int l=0,r=k-1;
while(l<=r)
{
int m=(r+l)>>1;
if(sum[m]+a[i]==tp)
{
//printf("%d\n",m);
flag=1;
break;
}
else if(sum[m]+a[i]>tp)
r=m-1;
else
l=m+1;
}
}
int main()
{
int l,n,m;
int cou=1;
while(~scanf("%d%d%d",&l,&n,&m))
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
memset(sum,0,sizeof(sum));
for(int i=0; i<l; i++)
scanf("%d",&a[i]);
for(int i=0; i<n; i++)
scanf("%d",&b[i]);
for(int j=0; j<m; j++)
scanf("%d",&c[j]);
k=0;
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
{
sum[k++]=b[i]+c[j];
}
sort(sum,sum+k);
int t;
scanf("%d",&t);
printf("Case %d:\n",cou++);
while(t--)
{
int tp;
scanf("%d",&tp);
flag=0;
for(int i=0; i<l; i++)
{
bs(tp,i);
}
if(!flag)
printf("NO\n");
else
printf("YES\n");
}
}
return 0 ;
}
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