project euler 20

Problem 20

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Factorial digit sum

n! means n × (n − 1) × … × 3 × 2 × 1

For example, 10! = 10 × 9 × … × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

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阶乘数字和

n! 的意思是 n × (n − 1) × … × 3 × 2 × 1

例如，10! = 10 × 9 × … × 3 × 2 × 1 = 3628800，所以10!的各位数字和是 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27。

求出100!的各位数字和。

package projecteuler;import java.util.ArrayList;import java.util.HashMap;import java.util.List;import java.util.Map;import java.util.Map.Entry;import org.junit.Test;public class Prj20 {/** * n! means n × (n − 1) × ... × 3 × 2 × 1 *  * For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the * digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. *  * Find the sum of the digits in the number 100! */@Testpublic void test() {System.out.println( Calculator.calculate(100));}public static class Calculator {/** * 1. 数组存取 2. 分解因子部分优化 2 * 5 = 10 , 不影响计算的去除 3. 乘法进一位 *  * @param n * @return */public static int calculate(int n) {int bits = getBits(n);int[] data = new int[bits];data[bits - 1] = 1; // initMap<Long, Integer> primeMap = new HashMap<Long, Integer>();IntegerDivisor div = new IntegerDivisor();for (int i = 2; i <= n; i++) {div.divisor(i);mergerPrimeMap(div.primeMap, primeMap);div.clear();}optimizeMap(primeMap);System.out.println( IntegerDivisor.print_Map(primeMap));for (Entry<Long, Integer> entry : primeMap.entrySet()) {Long prime = entry.getKey();int times = entry.getValue();for (int i = 1; i <= times; i++) {for (int bitIter = 0; bitIter < bits; bitIter++) {data[bitIter] *= prime;}for (int bitIter = bits - 1; bitIter > 0; bitIter--) {data[bitIter - 1] += (data[bitIter] / 10);data[bitIter] %= 10;}}}int sum = 0;for (int i = 0; i < bits; i++) {sum += data[i];System.out.println("data[" + i +"]=" + data[i]);}return sum;}private static void mergerPrimeMap(Map<Long, Integer> divPrimeMap,Map<Long, Integer> primeMap) {for( Entry<Long, Integer> entry : divPrimeMap.entrySet()){if( !primeMap.containsKey(entry.getKey())){primeMap.put( entry.getKey(), 0);}int time = primeMap.get( entry.getKey()) + entry.getValue();primeMap.put( entry.getKey(),  time);}}private static void optimizeMap(Map<Long, Integer> primeMap){if( primeMap.containsKey(2L) && primeMap.containsKey(5L)){int minSubtract = Math.min( primeMap.get(2L), primeMap.get(5L));int time2 = primeMap.get(2L);int time5 = primeMap.get(5L);primeMap.put((long) 2,  time2 - minSubtract);primeMap.put((long) 5,  time5 - minSubtract);if( primeMap.get(2L) == 0){primeMap.remove(2L);}if( primeMap.get(5L) == 0){primeMap.remove(5L);}}}public static int getBits(int n) {double bits = 0;for (int i = 1; i <= 100; i++) {bits += Math.log(i) / Math.log(2);}System.out.println(bits);return (int) (Math.ceil(bits) + 1);}}public static class IntegerDivisor {public Map<Long, Integer> primeMap = new HashMap<Long, Integer>();public List<Long> primeList = new ArrayList<Long>();public void clear() {primeMap.clear();primeList.clear();}public void divisor(long num) {if (num <= 1)return;long prime = getPrime(num,primeList.size() == 0 ? 2: primeList.get(primeList.size() - 1));if (prime < 0) {primeMap.put(num, 1);primeList.add(num);return;} else {primeList.add(prime);int count = 0;do {count += 1;num = num / prime;} while (num % prime == 0);primeMap.put(prime, count);divisor(num);}}private long getPrime(long num, long start) {for (long i = start; i <= Math.sqrt(num); i++) {if (num % i == 0) {return i;}}return -1;}@Overridepublic String toString() {return print_Map( this.primeMap);}public static String print_Map(Map<Long, Integer> primeMap) {StringBuilder sb = new StringBuilder();for (Entry<Long, Integer> entry : primeMap.entrySet()) {sb.append(entry.getKey().toString() + "="+ entry.getValue().toString() + "\n");}return sb.toString();}public Long getLargestPrime() {return primeList.get(primeList.size() - 1);}}}