連續取模-function
阿新 • • 發佈：2017-09-23
style mod calculate while present urn eno function case
You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r?1) modArl=r;l<r.F(l,r)={All=r;F(l,r?1) modArl<r.
You job is to calculate
2017-09-22 21:56:08
The shorter, the simpler. With this problem, you should be convinced of this truth.You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r?1) modArl=r;l<r.F(l,r)={All=r;F(l,r?1) modArl<r.
You job is to calculate
InputThere are multiple test cases.
The first line of input contains a integer TT, indicating number of test cases, and TT test cases follow.
For each test case, the first line contains an integer N(1≤N≤100000)N(1≤N≤100000).
The second line contains NN space-separated positive integers: A1,…,AN (0≤Ai≤109)A1,…,AN (0≤Ai≤109).
The third line contains an integer MM denoting the number of queries.
The following MM lines each contain two integers l,r (1≤l≤r≤N)l,r (1≤l≤r≤N), representing a query.
OutputFor each query(l,r)(l,r), output F(l,r)F(l,r) on one line.Sample Input
1
3
2 3 3
1
1 3
Sample Output
2
代碼如下：
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
#define MAXN 100010
int a[MAXN],nex[MAXN];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int i,j,n,m;
scanf("%d",&n);
for(i = 1; i<=n; ++i)
{
scanf("%d",&a[i]);
}
for(i = 1;i<=n;++i)
{
nex[i] = -1;
for(j = i+1;j<=n;++j)
{
if(a[j]<=a[i])
{
nex[i] = j;
break;
}
}
}
scanf("%d",&m);
for(i = 0;i<m;++i)
{
int l,r;
scanf("%d%d",&l,&r);
int num = a[l];
for(j = nex[l];j<=r;j = nex[j])
{
if(j == -1)
{
break;
}
num%=a[j];
}
printf("%d\n",num);
}
}
return 0;
}
連續取模-function