PHP classic question: Hundred Money and Hundred Chicken Problem (Exhaustive Algorithm)_PHP Tutorial

PHP classic question: Hundred Money and Hundred Chicken Problem (Exhaustive Algorithm)

Hundred dollars and hundred chickens problem:

Known: roosters are 5 yuan each, hens are 3 yuan each, and chicks are 3 yuan each

I bought 100 chickens with 100 yuan. Question: How many roosters, hens and chicks are there?

--Please consider the most efficient method possible

Things:

If there are 0 roosters, 0 hens, and 1 chick, is the number 100? Is the price 100? No

If there are 0 roosters, 0 hens, and 2 chicks, is the number 100? Is the price 100? No

If there are 0 roosters, 0 hens, and 3 chicks, is the number 100? Is the price 100? No

......

If there are 0 roosters, 0 hens, and 100 chicks, is the number 100? Is the price 100? No

If there are 0 roosters, 1 hen, and 1 chick, is the number 100? Is the price 100? No

If there are 0 roosters, 1 hen, and 2 chicks, is the number 100? Is the price 100? No

......

If there are 0 roosters, 1 hen, and 100 chicks, is the number 100? Is the price 100? No

If there are 0 roosters, 2 hens, and 1 chick, is the number 100? Is the price 100? No

......

If there are 100 roosters, 100 hens, and 0 chicks, is the number 100? Is the price 100? No

If there are 100 roosters, 100 hens, and 1 chick, is the quantity 100? Is the price 100? No

If there are 100 roosters, 100 hens, and 2 chicks, is the quantity 100? Is the price 100? No

......

This is called: exhaustive thinking (that is, testing all possible situations one by one)

PHP code:

echo "

Original idea:

";

$count = 0;

for($gongji = 0; $gongji <= 100; $gongji) {

for($muji = 0; $muji <= 100; $muji) {

for($xiaoji = 0; $xiaoji <= 100; $xiaoji) {

if($gongji*5 $muji*3 $xiaoji/3 == 100 && $gongji $muji $xiaoji == 100) {

echo "
There are $gongji roosters; $muji hens; and $xiaoji chicks;";

}

$count ; //Count times

}

}

}

echo "
Number of times: $count";

echo '
';

echo "

Code Optimization 1:

";

$count = 0;

for($gongji = 0; $gongji <= 100; $gongji) {

for($muji = 0; $muji <= 100; $muji) {

$xiaoji = 100 - $gongji - $muji;

if($gongji*5 $muji*3 $xiaoji/3 == 100) {

echo "
There are $gongji roosters; $muji hens; and $xiaoji chicks;";

}

$count ; //Count times

}

}

echo "
Number of times: $count";

echo '
';

echo "

Code Optimization 2:

";

$count = 0;

for($gongji = 0; $gongji <= 100/5; $gongji) { //Based on the total price: there are at most 100/5 roosters

for($muji = 0; $muji <= 100/3; $muji) { //Based on the total price: the maximum number of hens is 100/3

$xiaoji = 100 - $gongji - $muji;

if($gongji*5 $muji*3 $xiaoji/3 == 100) {

echo "
There are $gongji roosters; $muji hens; and $xiaoji chicks;";

}

$count ; //Count times

}

}

echo "
Number of times: $count";

echo '
';

echo "

Code optimization three:

";

$count = 0;

for($gongji = 0; $gongji <= 100/5; $gongji) { //Based on the total price: there are at most 100/5 roosters

for($muji = 0; $muji <= (100-$gongji*5)/3; $muji) { //Based on the total price and the money spent on the rooster: then the hen has at most (100- $gongji*5)/3 pieces

$xiaoji = 100 - $gongji - $muji;

if($gongji*5 $muji*3 $xiaoji/3 == 100) {

echo "
There are $gongji roosters; $muji hens; and $xiaoji chicks;";

}

$count ; //Count times

}

}

echo "
Number of times: $count";

echo '
';

echo "

Code optimization four:

";

$count = 0;

for($gongji = 0; $gongji <= 100/5; $gongji) { //Based on the total price: there are at most 100/5 roosters

for($muji = 0; $muji <= (100-$gongji*5)/3; $muji) { //Based on the total price and the money spent on the rooster: then the hen has at most (100- $gongji*5)/3 pieces

$xiaoji = 100 - $gongji - $muji;

if($xiaoji % 3 != 0) {continue;} //Considering the price of chickens, the number of chickens can only be reasonable if it is divisible by 3

if($gongji*5 $muji*3 $xiaoji/3 == 100) {

echo "
There are $gongji roosters; $muji hens; and $xiaoji chicks;";

}

$count ; //Count times

}

}

echo "
Number of times: $count";

Output results and calculation times:

Original idea:

There are 0 roosters; 25 hens; 75 chicks;

There are 4 roosters; 18 hens; 78 chicks;

There are 8 roosters; 11 hens; 81 chicks;

There are 12 roosters; 4 hens; 84 chicks;

Number of times: 1030301

Code Optimization 1:

There are 0 roosters; 25 hens; 75 chicks;

There are 4 roosters; 18 hens; 78 chicks;

There are 8 roosters; 11 hens; 81 chicks;

There are 12 roosters; 4 hens; 84 chicks;

Number of times: 10201

Code Optimization 2:

There are 0 roosters; 25 hens; 75 chicks;

There are 4 roosters; 18 hens; 78 chicks;

There are 8 roosters; 11 hens; 81 chicks;

There are 12 roosters; 4 hens; 84 chicks;

Number of times: 714

Code optimization three:

There are 0 roosters; 25 hens; 75 chicks;

There are 4 roosters; 18 hens; 78 chicks;

There are 8 roosters; 11 hens; 81 chicks;

There are 12 roosters; 4 hens; 84 chicks;

Number: 364

Code optimization four:

There are 0 roosters; 25 hens; 75 chicks;

There are 4 roosters; 18 hens; 78 chicks;

There are 8 roosters; 11 hens; 81 chicks;

There are 12 roosters; 4 hens; 84 chicks;

Number of times: 121

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