1007: Pass!
1 0
0 1
10 2
90 3
910 4
9090 5
90910 6
909090 7
9090910
90909090
909090910
\(f_i = f_{i-1} * 9 + f_{i-2} * 10\)
\(f_i + f_{i-1} = 10^x\)

\(if(奇) f_i = (f_{i-1} - 1) * 10\)
\(else f_i = (f_{i-1} + 1) * 10\)
1、假设x是偶数项 \(-> x + (x - 1) * 10 = 10^m\)\((m & 1 == 0)\)
2、假设x是奇数项 \(-> x + (x + 1) * 10 = 10^m\)\((m & 1 == 1)\)
通用即将10改为(n - 1)

TLE：
优化
1、偶数项优化：即 \(x\) + \((x - 1)\) * 10 = \(100^m\) -> \(ans = 2 * m\)
2、奇数项优化：变为 \(x\) * \(inv(n - 1) + 1 + x\) = \(100^m\) -> \(ans = 2 * m + 1\)

还是不行
优化：将两项放在一起 -> 显然两项都是\(100^m\)，一起求

还是不行
优化：将\(map\) -> \(unordered_map\)

ok!
BGSG:

const int mod = 998244353;
inline int ksm(int a, int b) {
    int res = 1;
    while(b) {
        if(b & 1)   res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
inline int inv(int x) {
    return ksm(x, mod - 2);
}

int log(int a, int b, int ans1, int ans2) {
    int m = sqrt(mod + 0.5), v = inv(ksm(a, m));
    unordered_map<int, int> mp;   mp[1] = 0;
    for(int i = 0, e = 1; i < m; ++ i, e = e * a % mod)  {
        if(e == ans1)   return i * 2;
        if(e == ans2)   return i * 2 + 1;
        if(!mp.count(e))    mp[e] = i;
    }
    for(int i = 0, mul = 1; i < m; ++ i, mul = mul * v % mod) {
        if(mp.count(ans1 * mul % mod))  return (i * m + mp[ans1 * mul % mod]) * 2;
        if(mp.count(ans2 * mul % mod))  return (i * m + mp[ans2 * mul % mod]) * 2 + 1;
    }
    return -1;
}
void run() {
    //printf("%lld\n", ksm(10, 196003445));
    int n = rd(), x = rd(), pre = 1;
    /*for(int i = 1; i <= 10; ++ i) {
        if(i & 1)   printf("%lld\n", pre = (pre - 1) * (n - 1));
        else        printf("%lld\n", pre = (pre + 1) * (n - 1));
    }*/
    if(x == 1)  {
        puts("0");
        return ;
    }
    if(x == 0) {
        puts("1");
        return ;
    }
    int y = (x - 1) * (n - 1) % mod;
    int ans1 = (x + y) % mod;
    //x = (x + 1) * (n - 1) % mod;
    //(prex - 1) * (n - 1) = x
    //prex = x / (n - 1) + 1
    y = (x * ksm(n - 1, mod - 2) + 1) % mod;
    int ans2 = (x + y) % mod;
    int res = log((n - 1) * (n - 1) % mod, (x + y) % mod, ans1, ans2);//偶数项
    printf("%lld\n", res % mod);
    /*x = (x + 1) * (n - 1) % mod;
    y = (x - 1) * (n - 1) % mod;
    res = log((n - 1) * (n - 1) % mod, (x + y) % mod);//奇数项
    if(res != -1) printf("%lld\n", (res * 2 % mod - 1 + mod) % mod);
    else puts("-1");*/
    return ;
}
signed main() {
    int t = rd();
    while(t--) run();
    return 0;
}

1010: zoto
\(x\)轴莫队，对\(y\)轴维护一个分块
这样修改的复杂度是\(O(1 * \sqrt n)\)的，查询的复杂度是$O(\sqrt n) $的
三维偏序 CDQ咋做呀

const int maxn = 1e5 + 10;
int n, m, sz, bnum;
int a[maxn], ans[maxn], pos[maxn];
struct query{
    int l, r, y1, y2, id;
    bool operator < (const query & a) const {
        return (pos[l] ^ pos[a.l]) ? pos[l] < pos[a.l] : ((pos[l] & 1) ? r < a.r : r > a.r);
    }
}q[maxn];

int num[maxn], sum[maxn];
inline void add(int x) {
    (!num[x]++)&&(++sum[x / sz]);
}
inline void del(int x) {
    (!--num[x])&&(--sum[x / sz]);
}
int calc(int x) {
    int res = 0;
    for(int i = 0; i < x / sz; ++ i)   res += sum[i];
    for(int i = x / sz * sz; i <= x; ++ i)  res += num[i] ? 1 : 0;
    return res;
}
void run() {
    n = rd(), m = rd();
    memset(num, 0, sizeof num);
    memset(sum, 0, sizeof sum);
    sz = sqrt(n), bnum = ceil(1.0 * n / sz);
    for(int i = 1; i <= bnum; ++ i) for(int j = (i - 1) * sz + 1; j <= i * sz; ++ j)    pos[j] = i;
    for(int i = 1; i <= n; ++ i)    a[i] = rd();
    for(int i = 1; i <= m; ++ i)
        q[i].l = rd(), q[i].y1 = rd(), q[i].r = rd(), q[i].y2 = rd(), q[i].id = i;
    sort(q + 1, q + 1 + m); int l = 1, r = 0;
    for(int i = 1; i <= m; ++ i) {
        while(l < q[i].l)   del(a[l++]);
        while(l > q[i].l)   add(a[--l]);
        while(r < q[i].r)   add(a[++r]);
        while(r > q[i].r)   del(a[r--]);
        ans[q[i].id] = calc(q[i].y2) - calc(q[i].y1 - 1);
    }
    for(int i = 1; i <= m; ++ i)    printf("%lld\n", ans[i]);
}