P2260 [清华集训2012]模积和
推导过程
我们假定
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
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 ,,,
代码
注意随手取模,容易溢出 。
/*
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;
}