P3396 哈希冲突(思维+方块)

题目

P3396 哈希冲突

做法

预处理模数\([1,\sqrt{n}]\)的内存池，\(O(n\sqrt{n})\)

查询模数在范围里则直接输出，否则模拟\(O(m\sqrt{n})\)

修改则遍历模数\([1,\sqrt{n}]\)，复杂度\(O(m\sqrt{n})\)

Code

#include<bits/stdc++.h>
typedef int LL;
const LL maxn=2e5+9;
inline LL Read(){
	LL x(0),f(1); char c=getchar();
	while(c<'0' || c>'9'){
		if(c=='-') f=-1; c=getchar();
	}
	while(c>='0' && c<='9'){
		x=(x<<3ll)+(x<<1ll)+c-'0'; c=getchar();
	}return x*f;
}
LL n,m;
LL a[maxn],sum[509][509];
int main(){
	n=Read(); m=Read();
	for(LL i=1;i<=n;++i) a[i]=Read();
	LL up(sqrt(n));
	for(LL i=1;i<=n;++i)
		for(LL j=1;j<=up;++j) sum[j][i%j]+=a[i];
	while(m--){
		char c; scanf(" %c",&c);
		LL x(Read()),y(Read());
		if(c=='A'){
			if(x<=up) printf("%d\n",sum[x][y]);
			else{
				LL ret(0);
				for(LL i=y;i<=n;i+=x) ret+=a[i];
				printf("%d\n",ret);
			}
		}else{
			for(LL j=1;j<=up;++j) sum[j][x%j]-=a[x],sum[j][x%j]+=y;
			a[x]=y;
		}
	}
	return 0;
}