Parco_Love_String
https://ac.nowcoder.com/acm/contest/625/K
思路:字符哈希(get),先记录全部区间的串,再枚举切点,每次把新的子串加到左边,同时总的减去切点为l的,因为之后就没这个区间的事了,还有在sum中减去重复计算的;
直接map会超时,unorder_map(get)
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x) memset(x,0,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e3+9;
const int mod=1e9+7;
template <class T>
inline void sc(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
ull seed=20123;
char s[maxn];
ull has[maxn];
ull base[maxn];
ll ans[maxn];
unordered_map<ull,int>mptol,mpl;
//map<ll,int>mptol,mpl;
ull string_hash(int l,int r)
{
return has[r]-has[l-1]*base[r-l+1];
}
int main()
{
FAST_IO;
//freopen("input.txt","r",stdin);
base[0]=1;
scanf("%s",s+1);
//printf("%s\n",s+1);
int len=strlen(s+1);
for(int i=1;i<=len;i++) base[i]=base[i-1]*seed;
has[0]=0;
for(int i=1;i<=len;i++) has[i]=has[i-1]*seed+s[i]-'a'+1;//要加1!!!
for(int i=1;i<=len;i++)
{
for(int j=i;j<=len;j++) mptol[(string_hash(i,j))]++;
}
ll sum=0;
for(int i=1;i<len;i++)
{
for(int j=i;j>=1;j--)
{
//cout<<mptol[string_hash(j,i)]<<" ";
mpl[string_hash(j,i)]++;
if(mptol[string_hash(j,i)]) sum+=mptol[string_hash(j,i)];
}
//cout<<sum<<" ";
for(int j=i;j<=len;j++)
{
mptol[string_hash(i,j)]--;
if(mptol[string_hash(i,j)]==0) mptol.erase(string_hash(i,j));
if(mpl[string_hash(i,j)]) sum-=mpl[string_hash(i,j)];
}
//cout<<sum<<endl;
ans[i]=sum;
}
int t;
//cin>>t;
sc(t);
while(t--)
{
int id;
//cin>>id;
//cout<<ans[id]<<endl;
sc(id);
printf("%lld\n",ans[id]);
}
return 0;
}