HDU - 5875 Function(预处理)
Function
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 F(l,r)F(l,r), for each query (l,r)(l,r).
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
预处理出每个数下一个比他小(或等于)的数的位置。然后跳着取模即可。
数据是有多水。。n^2预处理都能过。。
#include <bits/stdc++.h>
#define MAXN 100005
using namespace std; int a[MAXN],nxt[MAXN]; int main(void)
{
int T,n,m,l,r,ans;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
for(int i = ; i <= n; i++) {
scanf("%d",&a[i]);
nxt[i]=n+;
}
for(int i=;i<=n;i++){
for(int j=i+;j<=n;j++){
if(a[i]>=a[j]){
nxt[i]=j;
break;
}
}
}
scanf("%d",&m);
while(m--) {
scanf("%d %d",&l,&r);
int ans=a[l];
for(int i=nxt[l];i<=r;i=nxt[i]){
if(i==n+) break;
ans%=a[i];
}
printf("%d\n",ans);
}
}
return ;
}
暴力预处理(非正解)
#include <bits/stdc++.h>
#define MAXN 100005
#define lson num << 1
#define rson num << 1 | 1
using namespace std;
struct node
{
int l,r;
int Min;
}tree[MAXN << ];
int a[MAXN],d1[MAXN];
int cur;
void pushup(int num)
{
tree[num].Min = min(tree[lson].Min,tree[rson].Min);
}
void build(int num,int l,int r)
{
tree[num].l = l;
tree[num].r = r;
if(l == r) {
tree[num].Min = a[l];
return;
}
int mid = (l + r) >> ;
build(lson,l,mid);
build(rson,mid + ,r);
pushup(num);
}
void query1(int num,int l,int r,int val)
{
if(tree[num].l == tree[num].r) {
if(tree[num].Min <= val) cur = min(cur,tree[num].l);
return;
}
int mid = (tree[num].l + tree[num].r) >> ;
if(tree[num].l == l && tree[num].r == r) {
if(tree[lson].Min <= val) query1(lson,l,mid,val);
else if(tree[rson].Min <= val) query1(rson,mid + ,r,val);
return;
}
if(r <= mid) query1(lson,l,r,val);
else if(l > mid) query1(rson,l,r,val);
else {
query1(lson,l,mid,val);
query1(rson,mid + ,r,val);
}
}
int main(void)
{
int T,n,m,l,r,Max,pos,val,ans;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
Max = ;
for(int i = ; i <= n; i++) {
scanf("%d",&a[i]);
d1[i] = ;
}
build(,,n);
d1[n] = n + ;
for(int i = ; i <= n - ; i++) {
cur = n + ;
query1(,i + ,n,a[i]);
d1[i] = cur;
}
/*for(int i = 1; i <= n; i++) {
printf("%d--\n",d1[i]);
}*/
scanf("%d",&m);
while(m--) {
scanf("%d %d",&l,&r);
int ans=a[l];
for(int i=d1[l];i<=r;i=d1[i]){
if(i==n+) break;
ans%=a[i];
}
printf("%d\n",ans);
}
}
return ;
}
/*
10
5
5 8 10 3 2 */
线段树预处理(正解)
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