上海大都会 H.A Simple Problem with Integers

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题目描述

You have N integers A1, A2, ... , AN. You are asked to write a program to receive and execute two kinds of instructions:

1. C a b means performing Ai = (Ai2 mod 2018) for all Ai such that a ≤ i ≤ b.
2. Q a b means query the sum of Aa, Aa+1, ..., Ab. Note that the sum is not taken modulo 2018.

输入描述:

The first line of the input is T(1≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each test case contains N (1 ≤ N ≤ 50000).The second line contains N numbers, the initial values of A1, A2, ..., An.  0 ≤ Ai < 2018. The third line contains the number of operations Q (0 ≤ Q ≤ 50000). The following Q lines represents an operation having the format "C a b" or "Q a b", which has been described above. 1 ≤ a ≤ b ≤ N.

输出描述:

For each test case, print a line "Case #t:" (without quotes, t means the index of the test case) at the beginning.
You need to answer all Q commands in order. One answer in a line.

分析

刚开始打表算错了周期。。。又坑队友了
每个元素平方最大周期为6，且6是其他所有周期的公倍数。我们可以在每个节点维护一个大小为6的数组，同时维护一个代表从数组中取哪个元素的指针。因为有的数在进入循环节前需要经过几次修改，打表发现进入循环节前的修改次数都不超过5，所以可以在前五次暴力更新节点，之后才开始利用线段树的lazy标记。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod=2018;
const int maxn=50050;
int cnt[maxn*4],sum[maxn*4][7],lazy[maxn*4],p[maxn*4];
void build(int l,int r,int id) {
    cnt[id]=p[id]=lazy[id]=0;
    if(l==r) {
        scanf("%d", &sum[id][0]);
    }
    else {
        int mid=(l+r)>>1;
        build(l,mid,id<<1);
        build(mid+1,r,id<<1|1);
        sum[id][0]=sum[id<<1][0]+sum[id<<1|1][0];
    }
}
void pushUp(int id) {
    cnt[id]=min(cnt[id<<1],cnt[id<<1|1]);
    if(cnt[id]>=5) {
        p[id]=0;
        for (int i = 0; i < 6; ++i)
            sum[id][i]=sum[id<<1][(p[id<<1]+i)%6]+sum[id<<1|1][(p[id<<1|1]+i)%6];
    }
    else sum[id][0]=sum[id<<1][p[id<<1]]+sum[id<<1|1][p[id<<1|1]];
}
void pushDown(int id) {
    p[id<<1]=(p[id<<1]+lazy[id])%6,lazy[id<<1]+=lazy[id];
    p[id<<1|1]=(p[id<<1|1]+lazy[id])%6,lazy[id<<1|1]+=lazy[id];
    lazy[id]=0;
}
void modify(int x,int y,int l,int r,int id) {
    if(l>r||l>y||r<x) return;
    if(l==r) {
        cnt[id]++;
        if(cnt[id]<5) {
            sum[id][0]=sum[id][0]*sum[id][0]%mod;
        }
        else if(cnt[id]==5) {
            p[id]=0;
            sum[id][0]=sum[id][0]*sum[id][0]%mod;
            for (int i = 1; i < 6; ++i)
            {
                sum[id][i]=sum[id][i-1]*sum[id][i-1]%mod;
            }
        }
        else {
            p[id]=(p[id]+1)%6;
        }
        return;
    }
    if(lazy[id]) pushDown(id);
    if(x<=l&&y>=r) {
        int mid=(l+r)>>1;
        if(cnt[id]<5) {
            modify(x,y,l,mid,id<<1);
            modify(x,y,mid+1,r,id<<1|1);
            pushUp(id);
        }
        else {
            lazy[id]++;
            p[id]=(p[id]+1)%6;
        }
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid) modify(x,y,l,mid,id<<1);
    if(y>mid) modify(x,y,mid+1,r,id<<1|1);
    pushUp(id);
}
int query(int x,int y,int l,int r,int id) {
    if(l!=r && lazy[id]) pushDown(id);
    if(l>=x&&r<=y) {
        return sum[id][p[id]%6];
    }
    int mid=(l+r)>>1;
    int ans=0;
    if(x<=mid) ans+=query(x,y,l,mid,id<<1);
    if(y>mid) ans+=query(x,y,mid+1,r,id<<1|1);
    return ans;
}
int main(int argc, char const *argv[])
{
    int t,Case=0;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        build(1,n,1);
        int q;
        scanf("%d", &q);
        printf("Case #%d:\n", ++Case);
        while(q--)
        {
            char str[3];
            int l,r;
            scanf("%s%d%d", str,&l,&r);
            if(str[0]=='Q') {
                printf("%d\n", query(l,r,1,n,1));
            }
            else {
                modify(l,r,1,n,1);
            }
        }
    }
    return 0;
}