NOJ——1649Find Sum（二分查找）

最新推荐文章于 2022-08-03 22:06:32 发布

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最新推荐文章于 2022-08-03 22:06:32 发布

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分类专栏：算法 NBUT

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本文介绍了一种利用二分查找算法解决给定序列中特定元素和的高效方法，通过实例演示了如何在有限时间内快速找到满足条件的元素组合。

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[1649] Find Sum
时间限制: 1000 ms　内存限制: 65535 K
问题描述
This problem is really boring.

You are given a number sequence S which contain n integers and m queries, each time i will give your two integers id1 and id2, you need to tell me whether i can get S[id1] + S[id2] from the n integers.
输入
Input starts with an integer T(T <= 5) denoting the number of test cases.
For each test case, n(1 <= n <= 100000, 1 <= m <= 10000) are given, and the second line contains n integers, each of them is larger then 0 and no larger than 10^11.
Next m lines, each line contains two integers id1 and id2(1 <= id1, id2 <= n).
输出
For each case, first print the case number.
For each query, print “Yes” if i can get S[id1] + S[id2] from S, otherwise print “No”(without the quote).
样例输入
```
1
5 3
3 4 2 1 3
1 2
3 4
4 5
```
样例输出
```
Case 1:
No
Yes
Yes
```

会了二分查找之后这类题型做起来简直爽翻。科科~果然二分一般都是最先接触的算法吗。。。注意一点就是题目中给的数据或者说中间数据会超出int范围，第一次交RE了，改成__int64就AC了

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
bool bi_search(const __int64 list[],const __int64 &len,const __int64 &goal)
{
	__int64 l=0,r=len-1;
	while (l<=r)
	{
		__int64 mid=(l+r)>>1;
		if(list[mid]==goal)
			return true;
		else if(list[mid]<goal)
		{
			l=mid+1;
		}	
		else
		{
			r=mid-1;
		}
	}
	return false;
}
int main(void)
{
	int t,q;
	scanf("%d",&t);
	for (q=1; q<=t; q++)
	{
		__int64 n,m,i,d1,d2;
		scanf("%I64d%I64d",&n,&m);
		__int64 *list=new __int64[n];//用来放初始值
		__int64 *tlist=new __int64[n];//用来保存sort之后的有序序列，不然无法二分
		for (i=0; i<n; i++)
		{
			scanf("%I64d",&list[i]);
			tlist[i]=list[i];
		}	
		sort(tlist,tlist+n);
		printf("Case %d:\n",q);
		for (i=0; i<m; i++)
		{
			scanf("%I64d %I64d",&d1,&d2);
			if(bi_search(tlist,n,list[d1-1]+list[d2-1]))
				printf("Yes\n");
			else
				printf("No\n");
		}
		delete []list;
		delete []tlist;
	}
	return 0;
}