c Algorithm master, please enter 20

It seems that there are infinite numbers of such sequences.

For a fixed-number sequence, its composition is infinite and uncountable (since the members of the sequence can be any real number in [1,50], which is dense and uncountable). So, your last request is unattainable.

1l thinks too much, his meaning is obviously, take a few integers, so that their sum is 100, so it can still be achieved, it is recommended to use the recursive algorithm, from 1-50 in turn, if the sum is 100, then the output, less than, then continue to recursion, greater than, then exit this loop. I won't give it to you, it's time-consuming.

The basic idea is that for the array arr[50], first remove a number arr[k] from the array (k from 0 to 49);

In the remaining numbers, let i=k+1, j=max, if arr[i]+arr[j]=100-arr[k], then output these three numbers, i++, j--, enter the next round of cycle;

If arr[i]+arr[j]<100-arr[k], then i++;

If arr[i]+arr[j]>100-arr[k], then j--;

The outer time complexity is n, the inner layer is n, and the complexity of the whole algorithm is n 2

for(k=0;k<50;k++)

sum = 100 - arr[k];

for(i=k+1,j=49;ielse j--;

The numeric subscript starts at 0, array[0....5,0...7] 1) The volume of array a = 6 (rows) 8 (columns) 6 (bytes) = 288 (bytes) 2) The address of the first byte of the last element of array a, a57:

loc(5,7)=loc(0,0)+5 8+7) 6 = 1000 (base address) + 282 = 1282

loc( 1,4 ) = loc( 0,0 ) 1×8+4 )×6 = 1000 + 72 = 1072

loc( 4,7 ) = loc( 0,0 ) 7×6+4 )×6 = 1000 + 276 = 1276

The first one, in fact, is a matrix multiplication, nesting of two cycles outside the curly braces n times, the degree is nxn=n squared, and then look at the curly braces, the inner cycle is n times, and the assignment is counted 1 time, but it is absorbed by n (n->infinity), so the whole number is n to the third power, and the algorithm time complexity is o (n to the third power).

Second, since each loop i is multiplied by 2, let the k-th cycle exit, that is, the k-power of 2 is greater than n, so k=log2n, where 2 is lower than logarithm, so the time complexity is o(log2n), also written o(logn), the difference is only the coefficient, this coefficient is ignored.

Expression evaluation rules.

n Expressions in parentheses are evaluated first.

The n operator is divided into 16 priorities, and the operators of different levels are adjacent to each other, and the operation with the highest priority is calculated first.

n sibling operators are adjacent and are calculated from left to right or right to left according to the associative law.

The order in which the components of each operation of the binary and ternary dry rock operators are calculated depends on the implementation.

#include /*1 */

int x,y ; 2 */

int f ( int z ){3 * no key royal x=x+1 ; 4 */

return (z*z); 5 */

void main ( 7 */

x=0 ; 8 */

printf ( d", x+f(2) ) ) bright friend * 9 * n calculation order.

From left to right, the result is 4

From right to left, the result is 5

Let u=20, v=30, w=40, x=20, y=50.

w==0 ||u+20>v ||xx

false ||true

xxtrue &&false

A lot of vocabulary It is recommended that you buy a book I shoot Sun You If you want to contact the algorithm There are many binary, occimal, and hexadecimal attack chains It depends on how you make up the answer order.

It's an algorithm similar to binary.

The algorithm is the solution to the problem as mentioned in our university textbooks. C++ is just a language tool. It's not useful to have a tool to make a good program, it's important to use the tool.

A computer algorithm is a detailed analysis of the process in which we use computer language to solve a practical problem!

a0=d

a1=d(1+r)-p

a2=(d(1+r)-p)(1+r)-p

d(1+r)^2-p(1+r)-p

a3=(d(1+r)^2-p(1+r)-p)(1+r)-p=d(1+r)^3-p(1+r)^2-p(1+r)-p>>>

am=d(1+r)^m-p(1+r)^(m-1)..p(1+r)-p

am<0

m=lgp-lg(p-d*r) lg(1+r) The conversion process is more troublesome, and it is estimated that you don't understand it, so forget it, if you really want to know, it depends on how the logarithm is simplified.

Solution: The meaning of a loan is compound interest.

d（1+r）^m=p（1+r）*1+p(1+r)^2+。。p(1+r)^m

So both sides begged for LG

lg（d）+m*lg(1+r)=lgp+lg((1+r)((1+r)^m-1)/(1+r-1))

Europe.. Did you write the formula wrong???