shift and/shift or算法

1. CF 914F Substrings in a String

大意: 给定一个串s, q个询问, (1)单点修改, (2)询问[l,r]范围内串y的出现次数.

shift and算法板子题

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e5+10;
#else
const int N = 11;
#endif

int n, m;
char s[N], buf[N];
bitset<N> ch[26], ans;

int main() {
	scanf("%s", s);
	n = strlen(s);
	REP(i,0,n-1) ch[s[i]-'a'].set(i);
	scanf("%d", &m);
	REP(i,1,m) {
		int op, x, y;
		char c;
		scanf("%d%d", &op, &x),--x;
		if (op==1) {
			scanf(" %c", &c);
			ch[s[x]-'a'].reset(x);
			s[x] = c;
			ch[s[x]-'a'].set(x);
		}
		else {
			scanf("%d%s", &y, buf),--y;
			int len = strlen(buf);
			if (y-x+1<len) {
				puts("0");
				continue;
			}
			ans.set();
			REP(i,0,len-1) ans&=ch[buf[i]-'a']>>i;
			int ret = (ans>>x).count()-(ans>>y-len+2).count();
			printf("%d\n", ret);
		}
	}
}