标签：二分 c++ stl set
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2141
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 11577 Accepted Submission(s): 3031
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
Author
wangye
三个数能否相加得到X; 把头两个相加,存入set. 然后X减去第三个数,一个个减过来. 然后用find在set 中二分来找,找到即是可以相加得到X;
之前全开__int64 MLE 了
开了两个set来分别存 结果TLE了
最后这样都化成int用一个set终于过了.
#include<stdio.h>
#include<set>
using namespace std;
#define ll __int64
int a[600];
int c[600];
int main()
{
int i,l,m,r,flag,op,tem,j;
set<int>my;
int cas=1;
while(scanf("%d%d%d",&l,&m,&r)!=EOF)
{
my.clear();
for(i=0;i<l;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<m;i++)
{
scanf("%d",&tem);
for(j=0;j<l;j++)
{
if((ll)tem+a[j]<=2147483647)
my.insert(tem+a[j]);
}
}
for(i=0;i<r;i++)
{
scanf("%d",&c[i]);
}
scanf("%d",&op);
printf("Case %d:\n",cas++);
while(op--)
{
scanf("%d",&tem);
flag=0;
for(i=0;i<r;i++)
{
if(my.find(tem-c[i])!=my.end())
{
flag=1;
break;
}
}
if(flag)
puts("YES");
else
puts("NO");
}
}
return 0;
}
//2147483647
/*
1 2 1
2147483647
1 0
0
3
*/
hdu 2141 Can you find it? 二分
标签：二分 c++ stl set
原文地址：http://blog.csdn.net/u013532224/article/details/39052179
