codeforces 797e

考虑$n,q leq1e5$,可以考虑使用根号算法在$O(sqrt n)$的时间内完成单次询问.

所以将询问分类讨论,当$k leq sqrt n$时,对于每一个k暴力DP求解.当$k geq sqrt n$时,直接暴力求解即可.

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#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<climits>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define inf 0x3f3f3f3f
#define N 100010
#define M
using namespace std;
inline int (){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
struct Query{
int q,k,ans,ord;
}q[N];
int n,m,a[N],f[N];
bool cmp(Query A,Query B){
return A.k<B.k;
}
bool recmp(Query A,Query B){
return A.ord<B.ord;
}
void dp(int k){
per(i,n,1){
if(i+a[i]+k<=n){
f[i]=f[i+a[i]+k]+1;
}else{
f[i]=1;
}
}
}
int main(){
n=read();
rep(i,1,n) a[i]=read();
m=read();
rep(i,1,m) q[i].q=read(),q[i].k=read(),q[i].ord=i;
sort(q+1,q+1+m,cmp);
rep(i,1,m){
if(q[i].k>300){
int ans=0,x=q[i].q;
while(x<=n){
x=x+a[x]+q[i].k;
ans++;
}
q[i].ans=ans;
}else{
if(q[i].k==q[i-1].k){
q[i].ans=f[q[i].q];
}else{
dp(q[i].k);
q[i].ans=f[q[i].q];
}
}
}
sort(q+1,q+1+m,recmp);
rep(i,1,m) printf("%dn",q[i].ans);
return 0;
}