bzoj 3207 可持久化线段树+hash

这道题要看出来这个做法还是比较容易说一下细节

1.因为要用hash的区间值域来建树，而hash为了不冲突要开的很大，所以值域就会比较的大，不过这道题好的一点是没有修改，所以直接离散一下就会小很多

2.hash的时候多mod (' '     )

3.mod 的值可以稍微取大一点

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 
 typedef long long ll;
 
 const ll maxn = ;
 const ll mod = (ll)1e14 + ;
 const ll p = ;
 
 ll a[maxn], b[maxn], pos[maxn], v[maxn], f[maxn], num = , n, m, k;
 
 ll int_get() {
     ll x = ; char c = (char)getchar(); bool f = ;
     while(!isdigit(c) && c != '-') c = (char)getchar();
     if(c == '-') c = (char)getchar(), f = ;
     while(isdigit(c)) {
         x = x *  + (int)(c - '');
         c = (char)getchar();
     }
     if(f) x = -x;
     return x;
 }
 
 struct node {
     ll data;
     node *l, *r;
 }e[maxn * ]; ll ne = ;
 node* root[maxn];
 
 void test(node* x, ll l, ll r) {
     if(!x) return;
     cout << l <<" "<< r << " "<< x-> data << endl;
     if(l ^ r) {
         ll mid = (l + r) >> ;
         test(x-> l, l, mid), test(x-> r, mid + , r);
     }
 }
 
 void insert(node* &x, node* y, ll l, ll r, ll v) {
     if(!x) x = e + ne ++;
     if(l == r) x-> data = (y ? y-> data : ) + ;
     else {
         ll mid = (l + r) >> ;
         if(v <= mid) {
             if(y) x-> r = y-> r, y = y-> l;
             insert(x-> l, y, l, mid, v);
         }
         else {
             if(y) x-> l = y-> l, y = y-> r;
             insert(x-> r, y, mid + , r, v);
         }
     }
 }
 
 ll ask(node* x, node* t, ll l, ll r, ll v) {
     if(!x) return ;
     if(l == r) return x-> data - (t ? t-> data : );
     else {
         ll mid = (l + r) >> ;
         if(v <= mid) {
             if(t) t = t-> l;
             return ask(x-> l, t, l, mid, v);
         }
         else {
             if(t) t = t-> r;
             return ask(x-> r, t, mid + , r, v);
         }
     }
 }
 
 bool cmp(ll x, ll t) {
     return b[x] < b[t];
 }
 
 void pre(ll n) {
     f[] = ;
     for(ll i = ; i <= n; ++ i) f[i] = f[i - ] * p % mod;
 }
 
 ll find(ll x) {
     ll l = , r = num + ;
     while(r - l > ) {
         ll mid = (l + r) >> ;
         if(b[pos[mid]] <= x) l = mid;
         else r = mid;
     }
     return (x == b[pos[l]] ? v[pos[l]] : -);
 }
 
 void read() {
     n = int_get(), m = int_get(), k = int_get();
     pre(n);
     for(ll i = ; i <= n; ++ i) a[i] = int_get();
     memset(b, , sizeof(b));
     for(ll i = ; i <= k; ++ i) b[] = (b[] * p % mod + a[i]) % mod;
     for(ll j = k + ; j <= n; ++ j)
         b[j - k + ] = (((b[j - k] - a[j - k] * f[k - ] % mod) % mod + mod) % mod * p % mod + a[j]) % mod;
     //for(ll i = 1; i <= n - k + 1; ++ i) cout << b[i] << endl;
     //cout << endl;
     for(int i = ; i <= n - k + ; ++ i) pos[i] = i;
     sort(pos + , pos +  + n - k + , cmp);
     v[pos[]] = ++ num;
     for(ll i = ; i <= n - k + ; ++ i) {
         if(b[pos[i]] != b[pos[i - ]]) ++ num;
         v[pos[i]] = num;
     }
     for(ll i = ; i <= n - k + ; ++ i) insert(root[i], root[i - ], , num, v[i]);
 }
 
 void sov() {
     while(m --) {
         ll ls, rs;
         ls = int_get(); rs = int_get(); rs -= k - ;
         ll x = ;
         for(ll i = ; i <= k; ++ i) {
             ll s = int_get();
             x = (x * p % mod + s ) % mod;
         }
         ll c = find(x);
         if(c == - || rs < ls) printf("Yes\n");
         else {
             if(ask(root[rs], root[ls - ], , num, c) > ) printf("No\n");
             else printf("Yes\n");
         }
     }
 }
 
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     read();
     //for(int i = 1; i <= n - k + 1; ++ i) test(root[i], 1, num), cout << endl;
     sov();
 }