P2260 [清华集训2012]模积和，P2834 能力测验（二维除法分块）

P2260 [清华集训2012]模积和

推导过程

我们假定 $n <= m$

$\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (n\mod i)(m \mod j), i \not= j$

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (n\mod i)(m \mod j) - \sum_{k = 1} ^{n} (n \mod k) (m \mod k)$

$= \sum_{i = 1} ^{n} (n - \lfloor \frac{n}{i}\rfloor i) \sum_{j = 1} ^{m} (m - \lfloor \frac{m}{j}\rfloor j) - \sum_{k = 1} ^{n} (n - \lfloor \frac{n}{k}\rfloor k) (m - \lfloor \frac{m}{k}\rfloor k)$

$= (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - \sum_{k = 1} ^{n}(nm - nk \lfloor{\frac{m}{k}}\rfloor - mk \lfloor{\frac{n}{k}}\rfloor) + k ^ 2 \lfloor{\frac{n}{k}}\frac{m}{k}\rfloor$

之后我们可以利用整除分块加平方和公式与逆元结合求得最后答案

$1 + 2 ^ 2 + 3 ^ 2 + …… + n ^ 2 = \frac{n(n + 1)(2n + 1)}{6}$

多余的化简，只要上面的就行了，一开始我化简成了下面的式子，以为更简单，结果求崩了，找不出 $bug$，，，

$= (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - n ^ 2m + \sum_{k = 1} ^{n}(nk \lfloor{\frac{m}{k}}\rfloor + mk \lfloor{\frac{n}{k}}\rfloor) - k ^ 2 \lfloor{\frac{n}{k}}\rfloor\lfloor\frac{m}{k}\rfloor$

$= (n ^ 2 - \sum_{i = 1} ^{n} \lfloor \frac{n}{i}\rfloor i)(m ^ 2 - \sum_{j = 1} ^{m} \lfloor \frac{m}{j}\rfloor j) - n ^ 2m + n ^ 2\sum_{k = 1} ^{n}k \lfloor{\frac{m}{k}}\rfloor + nm\sum_{k = 1} ^{n}k \lfloor{\frac{n}{k}}\rfloor) - \sum_{k = 1} ^{n}k ^ 2 \lfloor{\frac{n}{k}}\rfloor\lfloor\frac{m}{k}\rfloor$

代码

注意随手取模，容易溢出 $wa$。

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return x * f;
}

const int mod = 19940417, inv = 3323403;

ll calc1(ll n) {
    return n * (n + 1) % mod * (2 * n + 1) % mod * inv % mod;
}

ll calc2(ll l, ll r) {
    return (l + r) * (r - l + 1) / 2 % mod;
}

ll f(ll n) {
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans + n * (r - l + 1) % mod - calc2(l, r) * (n / l) % mod + mod) % mod;
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    ll n = read(), m = read();
    if(n > m) swap(n, m);
    ll ans = (f(n) * f(m)) % mod;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ll temp1 = n * m % mod * (r - l + 1) % mod;
        ll temp2 = n * calc2(l, r) % mod * (m / l) % mod;
        ll temp3 = m * calc2(l, r) % mod * (n / l) % mod;
        ll temp4 = (calc1(r) - calc1(l - 1) + mod) * (n / l) % mod * (m / l) % mod;
        ans = (ans - (temp1 - temp2 - temp3 + temp4) % mod + mod) % mod;
    }
    cout << ans << endl;
    return 0;
}

P2834 能力测验

照搬上面的代码，改一下模数和逆元即可。

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return x * f;
}

const int mod = 1000000007, inv = 166666668;

ll calc1(ll n) {
    return n * (n + 1) % mod * (2 * n + 1) % mod * inv % mod;
}

ll calc2(ll l, ll r) {
    return (l + r) * (r - l + 1) / 2 % mod;
}

ll f(ll n) {
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans + n * (r - l + 1) % mod - calc2(l, r) * (n / l) % mod + mod) % mod;
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    ll n = read(), m = read();
    if(n > m) swap(n, m);
    ll ans = (f(n) * f(m)) % mod;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ll temp1 = n * m % mod * (r - l + 1) % mod;
        ll temp2 = n * calc2(l, r) % mod * (m / l) % mod;
        ll temp3 = m * calc2(l, r) % mod * (n / l) % mod;
        ll temp4 = (calc1(r) - calc1(l - 1) + mod) * (n / l) % mod * (m / l) % mod;
        ans = (ans - (temp1 - temp2 - temp3 + temp4) % mod + mod) % mod;
    }
    cout << ans << endl;
    return 0;
}