P2260 [清華集訓2012]模積和

推導過程

我們假定$n<=m$

$$\sum _{i=1}^n\sum _{j=1}^m(n\mathrm{m}\mathrm{o}\mathrm{d}i)(m\mathrm{m}\mathrm{o}\mathrm{d}j),i=j$$

$$=\sum _{i=1}^n\sum _{j=1}^m(n\mathrm{m}\mathrm{o}\mathrm{d}i)(m\mathrm{m}\mathrm{o}\mathrm{d}j)?\sum _{k=1}^n(n\mathrm{m}\mathrm{o}\mathrm{d}k)(m\mathrm{m}\mathrm{o}\mathrm{d}k)$$

$$=\sum _{i=1}^n(n??\frac{n}{i}?i)\sum _{j=1}^m(m??\frac{m}{j}?j)?\sum _{k=1}^n(n??\frac{n}{k}?k)(m??\frac{m}{k}?k)$$

$$=(n^2?\sum _{i=1}^n?\frac{n}{i}?i)(m^2?\sum _{j=1}^m?\frac{m}{j}?j)?\sum _{k=1}^n(nm?nk?\frac{m}{k}??mk?\frac{n}{k}?)+k^2?\frac{n}{k}\frac{m}{k}?$$

之后我們可以利用整除分塊加平方和公式與逆元結合求得最后答案

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

多余的化簡，只要上面的就行了，一開始我化簡成了下面的式子，以為更簡單，結果求崩了，找不出$bug$，，，

$$=(n^2?\sum _{i=1}^n?\frac{n}{i}?i)(m^2?\sum _{j=1}^m?\frac{m}{j}?j)?n^{2}m+\sum _{k=1}^n(nk?\frac{m}{k}?+mk?\frac{n}{k}?)?k^2?\frac{n}{k}??\frac{m}{k}?$$

$$=(n^2?\sum _{i=1}^n?\frac{n}{i}?i)(m^2?\sum _{j=1}^m?\frac{m}{j}?j)?n^{2}m+n^2\sum _{k=1}^{n}k?\frac{m}{k}?+nm\sum _{k=1}^{n}k?\frac{n}{k}?)?\sum _{k=1}^n{k}^2?\frac{n}{k}??\frac{m}{k}?$$

代碼

注意隨手取模，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 | 1using 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 能力測驗

照搬上面的代碼，改一下模數和逆元即可。

/*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 | 1using 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; }

總結

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