hdu 5875 2016 ACM/ICPC Asia Regional Dalian Online 1008

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1427    Accepted Submission(s): 126


[align=left]Problem Description[/align]
The shorter, the simpler. With this problem, you should be convinced of this truth.

  

  You are given an array A
of N
postive integers, and M
queries in the form (l,r).
A function F(l,r) (1≤l≤r≤N)
is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.

You job is to calculate F(l,r),
for each query (l,r).
 

[align=left]Input[/align]
There are multiple test cases.

  

  The first line of input contains a integer T,
indicating number of test cases, and T
test cases follow.

  

  For each test case, the first line contains an integer
N(1≤N≤100000).

  The second line contains N
space-separated positive integers: A1,…,AN (0≤Ai≤109).

  The third line contains an integer M
denoting the number of queries.

  The following M
lines each contain two integers l,r (1≤l≤r≤N),
representing a query.
 

[align=left]Output[/align]
For each query(l,r),
output F(l,r)
on one line.
 

[align=left]Sample Input[/align]

1
3
2 3 3
1
1 3

 

[align=left]Sample Output[/align]

2

 

题意：

给出m个区间 （L,R），求出从L处的数左往右取模至R的答案；

思路：

用优先队列求出一个数向右的第一个比他小的值，在求答案的时候就可以直接跳转到比他小的值。

因为在和它取模之后，再和比它大的值取模是没有意义的。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxn =100005;
int data[maxn];
int flag[maxn];

priority_queue<int >q;

int main(){
int T;
scanf("%d",&T);
while(T--){
int n;
int i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&data[i]);

q.push(1);
for(i=2;i<=n;i++){
while(!q.empty()&&data[i]<data[q.top()]){
flag[q.top()]=i;
q.pop();
}
q.push(i);
}
while(!q.empty()){
flag[q.top()]=-1;
q.pop();
}

int m;
scanf("%d",&m);
for(i=1;i<=m;i++){
int l,r;
scanf("%d%d",&l,&r);
int fn=l;
int ans=data[l];
while(fn<=r){
int wtf=flag[fn];
if(wtf==-1)
break;
else if(wtf>r)
break;
else if(wtf<=r){
ans=ans%data[wtf];
fn=wtf;
}
if(ans==1)
break;
}
printf("%d\n",ans);
}

}
return 0;
}