POJ 1306 Combinations
作者：互联网
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
N things taken M at a time is C exactly.
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.Output
The output from this program should be in the form:N things taken M at a time is C exactly.
Sample Input
100 6 20 5 18 6 0 0
Sample Output
100 things taken 6 at a time is 1192052400 exactly. 20 things taken 5 at a time is 15504 exactly. 18 things taken 6 at a time is 18564 exactly.
读完题目发现 这是一道求排列组合的。但是公式已经给出。所以也就是求公式的值。
直接使用double是可以通过的。我这里给一个延展答案。就是如果数据量真的特别大大，double也放不下的那种长度的做法。
用string计算乘法，使用辗转相除法进行约分。
注释写得比较详细，请参考代码。
简答double 和复杂计算都有， 因为参加过一次考试，和这个题目比较类似。用的比较大的数据量计算。特别遗憾。在此记录
double
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class POJ1306{
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while (br.ready()) {
StringTokenizer st = new StringTokenizer(br.readLine());
double N = Integer.parseInt(st.nextToken());
double M = Integer.parseInt(st.nextToken());
if(N == 0 && M == 0) break;
double ans = 1;
int idx = 0;
for (double i = N; i > 0 ; i--) {
ans *= i;
idx++;
if(idx == M) break;
}
for (double i = M; i > 0 ; i--) {
ans /= i;
}
System.out.println(String.format("%.0f things taken %.0f at a time is %.0f exactly.", N, M, ans));
}
}
}
用string计算乘法, 超详细注释
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
// C = N! / (N-M)!M! 通过约分 == (N*N-1*...*N-M+1) / M!
public class VjudgeJ2 {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while (br.ready()) {
StringTokenizer st = new StringTokenizer(br.readLine());
int N = Integer.parseInt(st.nextToken()); // 总共的数据
int M = Integer.parseInt(st.nextToken()); // 挑选的数据
if(N == 0 && M == 0) break;
int[] fenzi = new int[N]; //公式的分子
int[] fenmu = new int[N]; //公式的分母
int idx = 0; // 存出分子分母数组的索引
for (int i = N; i > N - M ; i--) {
fenzi[idx++] = i;
}
idx = 0;
for (int i = M; i > 0 ; i--) {
fenmu[idx++] = i;
}
// 寻找最大公约数
while (true) {
boolean isEnd = true; // 不能继续进行约分
for (int i = 0; i < idx; i++) { // 遍历分子分母
for (int j = 0; j < idx; j++) {
int gdc = gdc(fenzi[i], fenmu[j]); // 辗转相除寻找最大公约数
if(gdc != 1) { // 找到的结果不等于1 可以进行约分
fenzi[i] /= gdc;
fenmu[j] /= gdc;
isEnd = false;
}
}
}
if(isEnd) break; // 当无法约分时跳出循环
}
String ans = "1";
for (int i = 0; i < idx; i++) {
ans = mult(fenzi[i]+"", ans); // 因为此题目特殊分母最后总是1，所以只需要相乘分子
}
// %d是数字 %s是字符串 %.0f是浮点型
System.out.println(String.format("%d things taken %d at a time is %s exactly.", N, M, ans));
}
}
// 辗转相除法。如果大数不能整除小数，则小数和余数的整除时，小数就是他们的最大公约数
private static int gdc(int a, int b) {
if(a > b){
int r = a % b;
if(r == 0) return b;
return gdc(b, r);
}
int r = b % a;
if(r == 0) return a;
return gdc(a, r);
}
// 大整数相乘
private static String mult(String a, String b) {
// 整数转换数组
char[] aarr = a.toCharArray();
char[] barr = b.toCharArray();
// 分别存储长度整数数组
char[] bigArr = aarr.length > barr.length ? aarr : barr;
char[] smallArr = aarr.length > barr.length ? barr : aarr;
// 模仿真正的乘法。midSum存储，相乘过程中的结果。
int[][] midSum = new int[smallArr.length][a.length()+b.length()];
// 模相乘过程中的结果进行叠加就是真正的值。
int[] sum = new int[a.length()+b.length()];
// 循环两个整数，从索引最大的开始，因为数字相乘从右到左进行
for (int i = smallArr.length-1; i >= 0; i--) {
int row = smallArr.length - 1 - i;
for (int j = bigArr.length - 1; j >= 0; j--) {
// 因为是char我们需要转换为int
int v = Integer.parseInt(smallArr[i]+"") * Integer.parseInt(bigArr[j]+"");
// 计算当相乘的结果应该放置过程数组的第几位。应该是两者所在位置相加
// 例如 10 * 10 计算时候 1*1 = 1 但是字符1的位置都是1，所以结果应该向前进两位变成100。
int col = bigArr.length - j - 1 + smallArr.length - i - 1;
midSum[row][col] += v; // 需要和本身位置的值进行相加，之前进过来的树需要累加
int tem = col;
// 如果结果大于9 需要继续向前进位。while循环进位。 因为有 ..9999这种情况
while (midSum[row][tem] > 9) {
int tv = midSum[row][tem];
midSum[row][tem] = tv % 10;
midSum[row][tem+1] += tv / 10;
tem+=1;
}
}
}
// 循环累加结果
for (int i = 0; i < bigArr.length + smallArr.length; i++) {
for (int j = 0; j < smallArr.length; j++) {
sum[i] += midSum[j][i]; // 我们需要行循环，列进行累加。所以内层循环用小数组。
int tem = sum[i];
int col = i;
// 累加 如果结果大于9 需要继续向前进位。while循环进位。 因为有 ..9999这种情况
while (tem > 9) {
sum[col] = tem % 10;
sum[col+1] += tem / 10;
tem = sum[col+1];
col+=1;
}
}
}
// 将数组位数的值转为字符串
StringBuilder sb = new StringBuilder();
boolean isT = false;
for (int i = bigArr.length + smallArr.length-1; i >=0; i--) {
if(sum[i] != 0) isT = true;
if(isT) sb.append(sum[i]);
}
return sb.toString();
}
}
标签：1306,tem,idx,int,smallArr,length,POJ,Combinations,new 来源： https://www.cnblogs.com/ives-xu/p/13744307.html