Function
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 976 Accepted Submission(s): 375Problem DescriptionThe 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).InputThere 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.OutputFor each query(l,r), output F(l,r) on one line.Sample Input1
3
2 3 3
1
1 3Sample Output2SourceRecommend
题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5875
题目大意：
N个数(N<=100000)，M个询问，每次询问L,R，求F(L,R)。
F(L,R)=F(L,R-1)%A[R] , L<R
=A[L] , L=R
题目思路：
【倍增】
这题数据水到暴力居然过了。。比赛的时候一想会T就没写。
首先这题实际是求A[L]%A[L+1]%A[L+2]...%A[R]的值。
那么很容易想到A[L]右边比他大的数都是没有必要取模的，变为找L右边第一个小于A[L]的取模
模完后的结果缩小，再次寻找当前位置右边第一个小于此时的值的继续取模。可以证明这样的取模次数为log2N(100000取模后最大为49999)
暴力的做法就是预处理的时候直接从后往前用单调栈求出位置X右边第一个小于A[X]的位置Y，next[X]=Y。查询的时候用next跳着模。
当然这是不够的，因为可以构造出递减序列使得实际操作次数仍为N。
用倍增的思想，对于位置i右侧的最长下降子序列，next[i][j]表示位置i右边第2j个的位置，预处理出每个位置的next数组
每次做的时候尽量远跳，直到找到next[j][0]恰好小于当前的值z，则z对A[next[j][0]]取模，移动左标记，继续查找，直到超出R或者没有更小的数。
倍增：
//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define N 100004
#define M 18
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
int a[N],s[N];
int nextt[N][M];
void init()
{
int i,j,k,top;
s[top=]=n;
for(i=;i<M;i++)nextt[n][i]=n+;
for(i=n-;i;i--)
{
while(top && a[i]<a[s[top]])top--;
if(!top)nextt[i][]=n+;
else nextt[i][]=s[top];
s[++top]=i;
}
for(j=;j<M;j++)
{
for(i=;i<n;i++)
{
k=nextt[i][j-];
nextt[i][j]=nextt[k][j-];
}
}
}
int work(int l,int r)
{
int i,j,z=a[l];
if(l==r)return z;
while(l<r)
{
if(!z)return z;
for(i=;i<M-;i++)
{
if(a[nextt[l][]]<=z || nextt[l][i]>r)break;
if(a[nextt[l][i+]]<=z || nextt[l][i+]>r)
{
l=nextt[l][i];
i=-;
continue;
}
}
if(i==M || nextt[l][i]>r || !a[nextt[l][i]])return z;
z%=a[nextt[l][]];l=nextt[l][];
}
return z;
}
int main()
{
#ifndef ONLINE_JUDGE
// freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
#endif
int i,j,k;
int x,y,z;
// init();
for(scanf("%d",&cass);cass;cass--)
// for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
// while(~scanf("%s",s))
// while(~scanf("%d",&n))
{
scanf("%d",&n);
for(i=;i<=n;i++)scanf("%d",a+i);
init();
scanf("%d",&m);
for(i=;i<=m;i++)
{
scanf("%d%d",&x,&y);
printf("%d\n",work(x,y));
}
}
return ;
}
/*
// //
*/
暴力：
//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define N 100004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
int a[N],pre[N],s[N];
void print()
{
int i;
for(i=;i<=n;i++)
printf("%d ",pre[i]);
puts("");
}
int main()
{
#ifndef ONLINE_JUDGE
// freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
#endif
int i,j,k;
int x,y,z,top,l,r;
// init();
for(scanf("%d",&cass);cass;cass--)
// for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
// while(~scanf("%s",s))
// while(~scanf("%d",&n))
{
scanf("%d",&n);
for(i=;i<=n;i++)
{
scanf("%d",&a[i]);
}
top=;
s[top]=n;pre[n]=n+;
for(i=n-;i;i--)
{
while(top && a[i]<=a[s[top]])
top--;
if(!top)pre[i]=n+;
else pre[i]=s[top];
s[++top]=i;
}
scanf("%d",&m);
for(i=;i<=m;i++)
{
scanf("%d%d",&l,&r);
x=a[l];
for(j=pre[l];j<=r;j=pre[j])
x%=a[j];
printf("%d\n",x);
}
//print();
}
return ;
}
/*
// //
*/