CodeForces - 1207F Remainder Problem(分块)

题目链接：点击查看

题目大意：给出一个长度为 500000500000500000 的数组，初始时全部为 000，需要执行 nnn 次操作，每次操作分为两种类型：

1xy1 \ x \ y1 x y：ax+=ya_x+=yax+=y
2xy2 \ x \ y2 x y：统计∑i∈R(x,y)ai\sum_{i \in R(x,y)}a_i∑i∈R(x,y)ai，其中 R(x,y)R(x,y)R(x,y) 是 [1,500000][1,500000][1,500000] 中对 xxx 取余等于 yyy 的位置

题目分析：考虑根号分块，即小范围暴力大范围通过预处理获得，当 x≤500000x \le \sqrt {500000}x≤500000 时，直接维护一个前缀和 sumi,jsum_{i,j}sumi,j 代表 x=i,y=jx = i,y = jx=i,y=j 时的答案；当 x≥500000x \ge \sqrt {500000}x≥500000 时，暴力查询的时间复杂度是 500000\sqrt {500000}500000

考虑修改，每次直接暴力修改 sumsumsum 数组即可，时间复杂度同样是 500000\sqrt {500000}500000 的，所以这个题分块实现两个操作的时间复杂度均摊下来就是 O(nn)O(n\sqrt n)O(nn) 的

代码：

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<unordered_map>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f;
}
template<typename T>
inline void write(T x)
{if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
LL sum[1010][1010],a[N];
int main()
{#ifndef ONLINE_JUDGE
//  freopen("data.in.txt","r",stdin);
//  freopen("data.out.txt","w",stdout);
#endif
//  ios::sync_with_stdio(false);int n;read(n);for(int i=1;i<=n;i++){int op,x,y;read(op),read(x),read(y);if(op==1){for(int i=1;i<=1000;i++) sum[i][x%i]+=y;a[x]+=y;}else{if(x<=1000) write(sum[x][y]),putchar('\n');else{LL ans=0;for(int i=y;i<=500000;i+=x) ans+=a[i];write(ans),putchar('\n');}}}return 0;
}

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