Function HDU - 5875
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:
You job is to calculate F(l,r), for each query (l,r)
.
Input
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.
Output
For each query(l,r), output F(l,r)
on one line.
Sample Input
1
3
2 3 3
1
1 3
Sample Output
2
题意:
给定区间l,r求上述公式
分析:
预处理出右边第一个比它小的数,然后询问的话就可以跳着向后找,直接暴力就可以
code:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000+7;
int pos[maxn];
int a[maxn],n,m;
void init(){
memset(pos,-1,sizeof(pos));
for(int i = n-1; i >= 1; i--){
int p = i + 1;
for(;;){
if(a[i] > a[p]){
pos[i] = p;
break;
}
if(pos[p] == -1){
pos[i] = -1;
break;
}
p = pos[p];
}
}
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
}
init();
scanf("%d",&m);
for(int i = 1; i <= m; i++){
int l,r;
scanf("%d%d",&l,&r);
int ans = a[l];
l = pos[l];
while(l <= r && l != -1){
ans %= a[l];
l = pos[l];
}
printf("%d\n",ans);
}
}
return 0;
}