Polynomial Division 数学题

https://www.hackerrank.com/contests/101hack45/challenges/polynomial-division

询问一个多项式能否整除一个一次函数。a * x + b

注意到如果能整除，就比如是x^2 + 2 * x + 1能整除2 * x + 2

那么它必定能整除2 * x + 2的根，也就是和根肯定有交点。

因为你能整除，也就是(x^2 + 2 * x + 1) = k * (2 * x + 2)

那么k * (2 * x + 2)还是条直线。唯独使得2 * x + 2 = 0那个点是不会变的。

然后就是bit维护了。相当于询问[L, R]中，这一段的和，

注意特判一下b = 0，有点不同。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int MOD = 1e9 + 7;
const int maxn = 1e5 + 20;
LL powx[maxn];
LL quick_pow(LL a, LL b, int MOD) {
    LL ans = 1;
    LL base = a;
    while (b > 0) {
        if (b & 1) {
            ans *= base;
            if (ans >= MOD) ans %= MOD;
        }
        b >>= 1;
        base *= base;
        if (base >= MOD) base %= MOD;
    }
    return ans;
}
LL c[maxn];
int n, a, b, q;
int lowbit(int x) {
    return x & (-x);
}
void upDate(int pos, LL val) {
    while (pos <= n) {
        c[pos] += val;
        pos += lowbit(pos);
        if (c[pos] >= MOD) c[pos] %= MOD;
    }
}
LL get_sum(int pos) {
    LL ans = 0;
    while (pos) {
        ans += c[pos];
        pos -= lowbit(pos);
        if (ans >= MOD) ans %= MOD;
    }
    return ans;
}
LL arr[maxn];
void work() {
//    cout << quick_pow(2, 4, MOD) << endl;
    scanf("%d%d%d%d", &n, &a, &b, &q);
    powx[0] = 1;
    powx[1] = -b * quick_pow(a, MOD - 2, MOD) % MOD;
    for (int i = 2; i <= n; ++i) {
        powx[i] = powx[i - 1] * powx[1] % MOD;
    }
    for (int i = 1; i <= n; ++i) {
        LL x;
        scanf("%lld", &x);
        arr[i] = x;
        upDate(i, x * powx[i - 1] % MOD);
    }
    if (b == 0) {
        while (q--) {
            int flag;
            scanf("%d", &flag);
            if (flag == 1) {
                int pos, val;
                scanf("%d%d", &pos, &val);
                ++pos;
                arr[pos] = val;
            } else {
                int L, R;
                scanf("%d%d", &L, &R);
                L++;
                R++;
                if (arr[L] == 0) {
                    printf("Yes
");
                } else printf("No
");
            }
        }
        return;
    }
    while (q--) {
        int flag;
        scanf("%d", &flag);
        if (flag == 1) {
            int pos;
            LL val;
            scanf("%d%lld", &pos, &val);
            pos++;
            LL now = (get_sum(pos) + MOD - get_sum(pos - 1)) % MOD;
            upDate(pos, -now);
            upDate(pos, val * powx[pos - 1] % MOD);
        } else {
            int L, R;
            scanf("%d%d", &L, &R);
            L++;
            R++;
            LL now = (get_sum(R) - get_sum(L - 1) + MOD) % MOD;
            if (now == 0) {
                printf("Yes
");
            } else printf("No
");
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}