2018 ICPC 焦作区域赛 Resistors in Parallel（找规律+大数）

传送门

题目大意

给出电阻的并联公式，规定一个含有平方因子的数的1R=0\frac{1}{R}=0R1=0。定义一个数的阻值为其所有的因子阻值并联求出的结果，问nnn以内并联后的最大的阻值是多少，输出分数形式

解题思路

打表不难知道在很大一段范围内答案是不变的，这体现在前几个数为1,2,6,30,210,2310...1,2,6,30,210,2310...1,2,6,30,210,2310...，不难发现恰好是1,1∗2,1∗2∗3,1∗2∗3∗5,1∗2∗3∗5∗7,1∗2∗3∗5∗7∗11...1,1*2,1*2*3,1*2*3*5,1*2*3*5*7,1*2*3*5*7*11...1,1∗2,1∗2∗3,1∗2∗3∗5,1∗2∗3∗5∗7,1∗2∗3∗5∗7∗11...

但是这个时候如何去找规律呢，我曾尝试质因数分解，然后递推去求，但是这样最后的因子个数会达到21002^{100}2100个，明显爆了。后来也想不出什么，去搜了题解，结果竟然是找了积性函数的规律。

设f(x)f(x)f(x)为xxx的因子并联的阻值，显然f(p)=11+1p=pp+1f(p)=\frac{1}{1+\frac{1}{p}}=\frac{p}{p+1}f(p)=1+p11=p+1p，而f(p1p2)=11+1p1+1p1p2+1p2=p1p2p1+2p1p2+p2=f(p1)f(p2)f(p_1p_2)=\frac{1}{1+\frac{1}{p_1}+\frac{1}{p_1p_2}+\frac{1}{p_2}}=\frac{p_1p_2}{p_1+2p_1p_2+p_2}=f(p_1)f(p_2)f(p1p2)=1+p11+p1p21+p211=p1+2p1p2+p2p1p2=f(p1)f(p2)

使用Java大数，那么这样就能愉快的预处理了！

import java.math.BigInteger;
import java.util.*;public class Main {static long prime[] = new long[1005];static boolean vis[] = new boolean[1005];static BigInteger fz[] = new BigInteger[1005], fm[] = new BigInteger[1005];static BigInteger array[] = new BigInteger[1005];static int cnt, T, tot;static void getPrime() {for (int i = 0; i < 1005; i++) vis[i] = false;vis[1] = false;cnt = 0;for (int i = 2; i < 1005; i++) {if (!vis[i]) {prime[++cnt] = i;}for (int j = 1; j <= cnt && i * prime[j] < 1005; j++) {vis[(int) (i * prime[j])] = true;if (i % prime[j] == 0) break;}}}static void init() {BigInteger Max = BigInteger.ONE;for (int i = 1; i <= 100; i++) {Max = Max.multiply(BigInteger.TEN);}BigInteger res = BigInteger.ONE;fz[0] = BigInteger.ONE;fm[0] = BigInteger.ONE;array[0] = BigInteger.ZERO;for (int i = 1; i <= cnt; i++) {res = res.multiply(BigInteger.valueOf(prime[i]));//System.out.println(res);array[i] = res;if (res.compareTo(Max) > 0) {break;}fm[i] = fm[i - 1].multiply(BigInteger.valueOf(prime[i]).add(BigInteger.ONE));fz[i] = fz[i - 1].multiply(BigInteger.valueOf(prime[i]));//System.out.println(fz[i] + "/" + fm[i]);BigInteger gcd = fz[i].gcd(fm[i]);fz[i] = fz[i].divide(gcd);fm[i] = fm[i].divide(gcd);}}public static void main(String[] args) {getPrime();init();BigInteger n;Scanner scanner = new Scanner(System.in);T = scanner.nextInt();while ((T--) != 0) {n = scanner.nextBigInteger();if (n.compareTo(BigInteger.ONE) == 0) {System.out.println("1/1");continue;}int idx = 0;for (int i = 1; ; i++) {if (n.compareTo(array[i]) == 0) {idx = i;break;}if (n.compareTo(array[i - 1]) > 0 && n.compareTo(array[i]) < 0) {idx = i - 1;break;}}if (fz[idx] == fm[idx]) {System.out.println("1/1");} else System.out.println(fz[idx] + "/" + fm[idx]);}}
}