NOJ——1649Find Sum（二分查找）

[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

代码：

#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:
",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
");
			else
				printf("No
");
		}
		delete []list;
		delete []tlist;
	}
	return 0;
}

NOJ——1649Find Sum（二分查找）

[1649] Find Sum