2017ICPC网络赛北京赛区 I题
2017ICPC网络赛北京赛区 I题
描述
You are given a list of integers a0, a1, …, a2^k-1.
You need to support two types of queries:
Output Minx,y∈[l,r] {ax∙ay}.
Let ax=y.
输入
The first line is an integer T, indicating the number of test cases. (1≤T≤10).
For each test case:
The first line contains an integer k (0 ≤ k ≤ 17).
The following line contains 2k integers, a0, a1, …, a2^k-1 (-2k ≤ ai < 2k).
The next line contains a integer (1 ≤ Q < 2k), indicating the number of queries. Then next Q lines, each line is one of:
1 l r: Output Minx,y∈[l,r]{ax∙ay}. (0 ≤ l ≤ r < 2k)
2 x y: Let ax=y. (0 ≤ x < 2k, -2k ≤ y < 2k)
输出
For each query 1, output a line contains an integer, indicating the answer.
样例输入
1
3
1 1 2 2 1 1 2 2
5
1 0 7
1 1 2
2 1 2
2 2 2
1 1 2
样例输出
1
1
4
题意：操作1求区间内最小的ax*ay，ax，ay可以相等。要注意ai的大小可以是负数，所以要进行分类讨论。当ai全部是负数的时候，答案是最大值x最大值；当ai全部是正数的时候，答案是最小值x最小值；当ai有正有负的时候，答案是最大值x最小值。
#include//prllf("%lld\n",n);
for(ll i=0;iscanf("%lld",&a[i]);
build(0,n-1,1);
ll q;
scanf("%lld",&q);
while(q--)
{
ll tag;
scanf("%lld",&tag);
if(tag==1)
{
ll x,y,ans;
scanf("%lld%lld",&x,&y);
ll minn=query1(x,y,1);
ll mann=query2(x,y,1);
//printf("%lld %lld\n",minn,mann);
if(mann<=0)
{
ans=mann*mann;
}
else if(minn>=0)
{
ans=minn*minn;
}
else
{
ans=mann*minn;
}
printf("%lld\n",ans);
}
else
{
ll x,y;
scanf("%lld%lld",&x,&y);
update(x,x,y,1);
}
}
}
return 0;
}