2

我想知道是否有任何人知道的一个公式可以预先计算循环中的迭代次数。

我玩弄了它并得出了等式：

迭代次数等于((终值与初值之差的绝对值)加(增/减的绝对值))除以增/减的绝对值

或者

预期 = (int)((Math.abs(final-initial) + Math.abs(iteration)) / 迭代),

例如：

for(int i = 43; i > 9; i-=8) ，将运行 5 次，因为 43-10 （如果不包括，则在初始值上加一）= 33+8 = 41/8 ~ 5

如果它等于或大于-for(int i = 43; I >= 9; i-=8)，那么根据我的公式你会做 (43-9+8) / 8 (5)

我已经测试了几次，但还不确定。

谢谢。

4

2 回答 2

1

排除任何特殊的循环中断（中断、更改计数器等），您可以测试以下公式。如果终止测试是<而不是<=你应该从开始减去 1。

total iterations = (end - start + incr)/incr; // for <=
total iterations = (end - start  + incr - 1)/incr; // for <

Random r = new Random();
for (int k = 0; k < 10000; k++) {
    int count = 0;
    int start = r.nextInt(20)+1;
    int end = r.nextInt(1000) + start+1;
    int incr = r.nextInt(50)+1;
    
    for (int i = start; i <= end; i += incr) {
        count++;
    }
    
    int calc = (end - start + incr - 1)/incr;
    if (calc != count) {
        System.out.println("Oops");
    }
}
于 2021-03-03T14:03:02.510 回答
0

排除任何特殊的循环中断（中断、更改计数器等），公式如下：

迭代次数 = (end - start+incr)/incr，其中关系运算符为 >= 或 <=

如果关系运算符是 < 或 >，则迭代次数 = (end - start+incr-1)/incr

这可以测试如下

/*
 * Baring any special disruption of the loop (breaks, altering counter, etc), the formula is as follows:

iterations = (|start-end|+|incr|)/|incr|, where relational operator is >= or <=

- if relational operator is < or >, then iterations = ((|start-end|+|incr|-1)/|incr|

 * @author rahul.arora, WJS
 */

import java.util.Random;

public class LoopIterationCalcTester {

    public static void main(String[] args) {
        
        
    for (int x = 0; x < 100; x++) {
        Random r = new Random();
        for (int k = 0; k < 10000; k++) {
            int count = 0;
            int start = r.nextInt(20)+1;
            int end = r.nextInt(1000) + 20;
            int incr = r.nextInt(50)+1;
            
            for (int i = start; i < end; i += incr) {
                count++;
            }
            
            int calc = (end - start-1+incr)/incr;
            
            if (calc != count) {
                System.out.println("Oops");
               
            }
        }
        
        
        for (int k = 0; k < 10000; k++) {
            int count = 0;
            int start = r.nextInt(20)+1;
            int end = r.nextInt(1000) + 20;
            int incr = r.nextInt(50)+1;
            
            for (int i = start; i <= end; i += incr) {
                count++;
            }
            
            int calc = (end - start+incr)/incr;
            
            if (calc != count) {
                System.out.println("Oops");
             
            }
        }
        for (int k = 0; k < 10000; k++) {
            int count = 0;
            int start = r.nextInt(20)+1;
            int end = r.nextInt(1000) + 20;
            int incr = r.nextInt(50)+1;
            
            for (int i = end; i >= start; i -= incr) {
                count++;
            }
            
            int calc = (end - start+incr)/incr;
            
            if (calc != count) {
                System.out.println("Oops");
            
            }
        }
        for (int k = 0; k < 10000; k++) {
            int count = 0;
            int start = r.nextInt(20)+1;
            int end = r.nextInt(1000) + 20;
            int incr = r.nextInt(50)+1;
            
            for (int i = end; i > start; i -= incr) {
                count++;
            }
            
            int calc = (end - start-1+incr)/incr;
            
            if (calc != count) {
                System.out.println("Oops");
            
            }
        }
        }
    }

}
于 2021-03-03T15:11:01.513 回答