gcd的性质+分块 Bzoj 4028

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespacestd;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")
/*首先，我们对所有的东西进行分块，然后我们对每个区间进行处理，分别得到这个区间的gcd和xor。
①然后如果前面的区间到这个块以后的gcd如果发生了改变，那么，我们就暴力这个块，复杂度为sqrt(n)
②如果前面的区间到这个块以后gcd没有发生改变，那么我们就二分这个块，二分的证明如下：
假定lastxor是该区间之前的xor值，lastgcd是该区间之前的gcd的值，假定我们要寻找的xor[j]是在这个块中
那么xor[j]^lastxor * lastgcd = k,转化以后为xor[j] = k/lastgcd ^ lastxor，所以我们只要二分这个块就好了
因此这里的复杂度为logn
因为gcd一共就只有logn个，所以复杂度最坏为sqrt(n) * logn个
*/
const int maxn = 1e6 + 5;
LL a[maxn], Xor[maxn], gcd[maxn];
intl[maxn], r[maxn], block, num, belong[maxn];
intn, q;
set<LL>S[maxn];

LL get_gcd(LL a, LL b){
    return b == 0 ? a : get_gcd(b, a %b);
}

voidbuild(){
    block = sqrt(n); num = n /block;
    if (n % block) num++;
    for (int i = 1; i <= num; i++)
        l[i] = (i - 1) * block + 1, r[i] = i *block;
    r[num] =n;
    for (int i = 1; i <= n; i++){
        belong[i] = (i - 1) / block + 1;
    }
}

void update(intp){
    S[p].clear();
    gcd[l[p]] = a[l[p]], Xor[l[p]] =a[l[p]];
    S[p].insert(Xor[l[p]]);
    for (int i = l[p] + 1; i <= r[p]; i++){
        gcd[i] = get_gcd(gcd[i - 1], a[i]);
        Xor[i] = Xor[i - 1] ^a[i];
        S[p].insert(Xor[i]);
    }
}

voidquery(LL val){
    LL prexor, pregcd;
    for (int i = 1; i <= r[1]; i++){
        if (gcd[i] * Xor[i] ==val){
            printf("%d
", i-1); return;
        }
    }
    prexor = Xor[r[1]], pregcd = gcd[r[1]];
    for (int i = 2; i <= num; i++){
        LL nowgcd =get_gcd(pregcd, gcd[r[i]]);
        if (nowgcd == pregcd){///xor[i] ^ prexor * nowgcd = val
            LL tmp = val / nowgcd ^prexor;
            if (val % nowgcd == 0 &&S[i].count(tmp)){
                for (int j = l[i]; j <= r[i]; j++){
                    if (Xor[j] ==tmp){
                        printf("%d
", j - 1); return;
                    }
                }
            }
        }
        else{
            for (int j = l[i]; j <= r[i]; j++){
                nowgcd =get_gcd(gcd[j], pregcd);
                LL tmp = val / nowgcd ^prexor;
                if (val % nowgcd == 0 && Xor[j] ==tmp){
                    printf("%d
", j - 1); return;
                }
            }
        }
        pregcd =get_gcd(pregcd, gcd[r[i]]);
        prexor = prexor ^Xor[r[i]];
    }
    printf("no
");
}

intmain(){
    cin >>n;
    for (int i = 1; i <= n; i++){
        scanf("%lld", a +i);
    }
    build();
    for (int i = 1; i <= num; i++)
        update(i);
    scanf("%d", &q);
    char ch[20];
    for (int i = 1; i <= q; i++){
        scanf("%s", ch);
        if (ch[0] == 'M'){
            intp; LL val;
            scanf("%d%lld", &p, &val);
            a[++p] =val;
            update(belong[p]);
        }
        else{
            LL val;
            scanf("%lld", &val);
            query(val);
        }
    }
    return 0;
}