// subject 49: The teacher divided the candy into several parts , Ask the students to collect it in any order , The first to receive , obtain 1 With the rest of the candy 1/10,
// The second one is , obtain 2 With the rest of the candy 1/10, The third one gets , obtain 3 With the rest of the candy 1/10, And so on ,
// How many students are there , How many sweets #include<stdio.h>
#include<stdlib.h> // Ideas : The requirement of this problem is to find a quantity of candy , This amount must ensure that each student receives the whole number of candy , Not all students are required to score the same candy
// Define variables according to multiple variables , Simplify the operation difficulty rule , Defined as x A student ,y Sweets
// By listing , Find out the rules :
//1+(y-1)*(1/10) The number of sweets the first student received
// Abstract functions
//f(1)=1+(y-1)*(1/10)
// Because every student gets the whole candy , therefore (y-1)*(1/10) Is an integer , explain y-1 yes 10 Multiple
// that y=10n+1
// By enumerating, we can get y The number of
// Define a variable last, To store the rest of the candy , Until last=0 until
// for the first time last=y- The number of sweets the first student received , namely last=y-1-(y-1)*(1/10) --- During enumeration y It's a known quantity
// The second time last=last- The number of sweets the second student received , namely last=last-(2+(last-2)*(1/10))
// You need to make sure that last-2 yes 10 Multiple , namely (last-2)%10==0, Don't meet this condition , The number of apples is about to change
//y=10n+1 take n++
// It also needs to be last>0
// When last==0 when , Get the normal number of students and candy
// When last<0 when , It means that the quantity of candy is not normal , Need to put n++ , Cycle again
// third time lats=last- The number of sweets the third student received , namely last=last-(3+(last-3)*(1/10)) // Uncertain number of cycles , Use while loop void main(){
// Define the self increasing variable
int n = ;
// Define the amount of candy left each time
int last = * n + ;
// Define the number of students
int index = ;
while (){
// When there's no cycle at all , The remaining candy is equal to the total candy
// Start the cycle
// For the first time
// The number of apples left must be 10 Multiple
if ((last - index)%!=)
{
// This shows that the total number of candy does not meet all the conditions
n++;
// Initialize all data
last = * n + ;
index = ;
// So let's do the next loop
continue;
}
last = last - (index + (last - index)/);
if (last == )
{
// It shows that the total quantity of candy meets all the conditions
break;
}
else if (last>)
{
index++;
}
else{
// This shows that the total number of candy does not meet all the conditions
n++;
// Initialize all data
last = * n + ;
index = ;
// So let's do the next loop
continue;
}
if (n>)
{
printf(" It seems that the code I wrote has gone wrong !\n");
break;
} }
if (index!=)
{
printf(" Altogether %d A student , There are a total of %d Share \n", index, * n + );
}
system("pause");
}

// subject 50: Kindergartens give candy to students from the front to the back , Each student gets candy in an arithmetic sequence , The sum of the first four students who got candy was 26, Product is 880,
// Seek before 20 The number of candies for each student #include<stdio.h>
#include<stdlib.h> // Ideas : Before finding out 20 A student's candy , You have to know the difference between a student's candy and the equal difference
//a1+a2+a3+a4=26, because a2=a1+d,a3=a1+(3-1)d; namely 4a1+d+2d+3d=26;=>4a1+6d=26
//a1*a2*a3*a4=880;a1*(a1+d)*(a1+2d)*(a1+3d)=880
// Candy is a positive integer , therefore 4a1+6d=26( namely 2a1+3d=13),a1 and d It's all integers , therefore a The range is 0<a1<7;0<d<5
// So use a double loop void main(){
int a[] = { };
int d = ;
for (int i = ; i < ; i++)
{
for (int j = ; j < ; j++)
{
if (*j + *i == )
{
if (j*(j + i)*(j + *i)*(j + *i) == )
{
a[] = j;
d = i;
break;
}
continue;
}
}
}
for (int i = ; i < ; i++)
{
if (i==)
{
printf("%5d", a[i]);
}
else{
a[i] = a[] + i*d;
printf("%5d", a[i]);
}
}
system("pause");
} // summary : This problem is an exercise, is to solve the binary equation programming solution , The key of binary equation programming is to determine the range of two elements , Find out the result by exhaustive method ,
// Good programming is to narrow down the scope of two elements , The scope of the problem element can be smaller ,2a1+3d=13,a1=6 In fact, it doesn't meet the requirements , Because at this time 3d=1,d It can't be an integer

// subject 51: Two different natural numbers A and B, If the whole number A All the factors of ( Include 1, barring A In itself ) The sum is equal to B;
// And integers B All the factors of ( Include 1, barring B In itself ) The sum is equal to A, Then the integer A and B It's called intimacy number . seek 10000 All intimacy numbers within . #include<stdio.h>
#include<stdlib.h>
#include<math.h> // Ideas : First, find a number A All the factors of , And then the sum of the factors , Get the number B, Number of judgements B Is the scope of 10000 within ,
// To find out B Is the sum of the factors equal to A
// Find all the factors of a number Find the number first A The square root of , And then we can find out all the factors by the method of circulation int run1(int num){
int a = (int)sqrt((double)num)+;
int sum = ;
for (int i = ; i < a; i++)
{
if (num%i==)
{
sum = sum + i + num / i;
//sum += i;
}
}
return sum;
} void main(){
int a[] = { };
int index = ;
int temp = ;
for (int i = ; i < ; i++)
{
// Acquisition number A The sum of the factors
temp = run1(i);
// Number of judgements B The scope of the
// Greater than required A, And in 10000 within , And count A It's not a number B
if (temp<i||temp>||(temp==i))
{
continue;
}
// Find the number B The sum of the factors
if (i == run1(temp))
{
printf("%d==>%d\n",i,temp);
}
}
system("pause");
} // summary : The difficulty of this problem is to find the number A Factor of , The key is to shrink A The range of factors (int)sqrt((double)num)+1

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