您的位置：首页 > 其它

HDOJ 2141 Can you find it?（二分搜索的优化+总结）

2017-08-11 21:21 441 查看

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 1328 Accepted Submission(s): 429

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

Sample Output

Case 1:
NO
YES
NO

Author

wangye

Source

HDU 2007-11 Programming Contest

Recommend

威士忌

直接用三重循环肯定超时。

只要把里面两重循环替换成二分搜索算法，就可以了。

先将里面两重循环的结果放到一个数组了。

然后把这个数组排序，因为二分查找需要该数组单调。

二分查找总结：

1.注意循环条件，直接循环100次的精度可以达到10的-30次方，如果控制成一定的精度退出循环，精度太小，由于浮点数偏差，可能陷入死循环。

2.二分查找要注意开区间和闭区间，如果（K】，找到的是第一个等于K的值。如果【K)找到的是最后一个等于K的值

3.二分适用于单调函数，三分适用于单峰函数

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int l, n, m;
int a[510];
int b[510];
int c[510];
int x[1010];
int lh[250010];
bool cmp(int a, int b)
{
return a < b;
}

bool binary(int x)
{

for (int i = 0; i < m; i++)
{
int first=0;
int last=n*l;
if (lh[first] + c[i] == x)return true;   //因为0的时候没有查找过
while (first < last - 1)
{
int mid = (first + last) / 2;
if (lh[mid] + c[i] == x)return true;
if (lh[mid] + c[i] > x)last = mid;
else first = mid;
}
}
return false;
}

int main()
{

int cnt = 0;
while (scanf("%d%d%d",&l,&n,&m)!=EOF)
{
cnt++;
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]);
}
for (int i = 0; i < l; i++)
{
for (int j = 0; j < n; j++)
{
lh[i*n + j] = a[i] + b[j];
}
}
sort(lh, lh + n*l, cmp);
int s;
scanf("%d", &s);
for (int i = 0; i < s; i++)
{
scanf("%d", &x[i]);
}
printf("Case %d:\n", cnt);
for (int i = 0; i < s; i++)
{

if(binary(x[i]))
printf("YES\n");
else
{
printf("NO\n");
}
}

}

return 0;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航