[Luogu] P3708 koishi的數學題
阿新 • • 發佈:2020-11-12
Description
輸入一個整數\(n\),設\(f(x) = \sum\limits_{i=1}^n x \bmod {i}\),你需要輸出\(f(1), f(2), \ldots , f(n)\)。\((n\le{10^6})\)
Solution
\(f(x)=\sum\limits_{i=1}^n x \bmod {i}=\sum\limits_{i=1}^n x-i\lfloor{\frac{x}{i}}\rfloor=nx-\sum\limits_{i=1}^ni\lfloor{\frac{x}{i}}\rfloor\)
\(f(x-1)=n(x-1)-\sum\limits_{i=1}^ni\lfloor{\frac{x-1}{i}}\rfloor\)
\(f(x)-f(x-1)=n-\sum\limits_{i=1}^ni(\lfloor{\frac{x}{i}}\rfloor-\lfloor{\frac{x-1}{i}}\rfloor)=n-\sum\limits_{i=1}^ni[i|x]=n-\sigma(x)\)
累加,有\(f(x)=nx-\sum\limits_{i=1}^{x}\sigma(i)\)
\(O(n)\)求出約數和即可。
(約數和及約數個數和見一個巨佬的部落格)
Code
#include <bits/stdc++.h> using namespace std; #define ll long long int n, tot, vis[1000005], pr[250005]; ll low[1000005], sd[1000005], sum[1000005]; int read() { int x = 0, fl = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') fl = -1; ch = getchar();} while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0'; ch = getchar();} return x * fl; } int main() { n = read(); sd[1] = 1; low[1] = 1; for (int i = 2; i <= n; i ++ ) { if (!vis[i]) { vis[i] = i; pr[ ++ tot] = i; low[i] = i + 1; sd[i] = i + 1; } for (int j = 1; j <= tot; j ++ ) { if (i * pr[j] > n || pr[j] > vis[i]) break; vis[i * pr[j]] = pr[j]; if (i % pr[j]) sd[i * pr[j]] = 1ll * sd[i] * (1ll + pr[j]), low[i * pr[j]] = 1ll + pr[j]; else sd[i * pr[j]] = 1ll * sd[i] / low[i] * (low[i] * pr[j] + 1ll), low[i * pr[j]] = 1ll * low[i] * pr[j] + 1ll; } } for (int i = 1; i <= n; i ++ ) sum[i] = sum[i - 1] + sd[i]; for (int i = 1; i <= n; i ++ ) printf("%lld ", 1ll * n * i - sum[i]); return 0; }