Determine whether each year in 2000-2500 is a leap year, and the results will be output:

 # include 
Int main () {

Int year; 
For (year=2000; Year & lt;=2500; Year++) {
If (year % 4==0 & amp; & Year % 100!=0 | | year %==0 400) {
Printf (" % d is a leap year \ n ", year); 
} 
The else {
Printf (" % d is not a leap year \ n ", year); 
} 
} 

return 0; 
} 

For 1 * 2 * 3 * 4 * 5:

 
#include 
Int main () {
Int t=1;//multiplier 
Int I=2;//the multiplicand 
While (I & lt; {
=5)T=t * I; 
i=i + 1; 
} 
Printf (" % d ", t); 
return 0; 
} 

There are 50 students to output results in more than 80 points students student number and grade:

 # include 
Int main () {
Int I, n [50], g [50].//I represents how many students are, n [I] on behalf of the student id, g [I] for achievement 
for(i=0; I & lt; 50; + + I) {
Printf (" please enter the first % d students student number and grade: ", I + 1); 
The scanf (" % d, % d ", & amp; N [I], & amp; G [I]); 
} 
for(i=0; I & lt; 50; + + I) {
[I] if (g & gt; {
=80)Student id: printf (" % d, achievement: % d \ n ", n [I], g [I]); 
} 
} 
return 0; 
} 

Given a positive integer greater than or equal to 3, whether it is a prime number:

 # include 
Int main () {
int n,i; 
Printf (" please enter a positive integer greater than or equal to 3: ", n); 
The scanf (" % d ", & amp; N); 
Int prime (int x); 
I=prime (n); 
If (I==1) {
Printf (" % d is not a prime \ n ", n); 
} else {
Printf (" % d is a prime \ n ", n); 
} 
return 0; 
} 
Int prime (int x) {
int y; 
If (x & gt; {
=3)For (y=2; Y & lt; X - 1; + + y) {
If (x % y==0) 
return 1; 
} 
return -1; 
} else {
Printf (" please enter a positive integer greater than or equal to 3! \n"); 
} 
} 

O 1-1/2 + a third - quarter +... + 1/99-1/100:

 # include 
Int main () {
Float term;//the value of the current item 
Float sum=1;//the current accumulation and 
Float deno=2;//the current item of the denominator 
Int sign=1; The current item number//symbol 
While (deno & lt; {
=100)Sign=(1) * sign; 
The term=sign * (1/deno); 
Sum=sum + term; 
Deno=deno + 1; 
} 
Printf (" % f ", sum); 
return 0; 
} 

CodePudding user response:

good

CodePudding user response:

Reference:

//given a positive integer greater than or equal to 3, whether it is a prime number: 

#include 
Int main () {

Int prime (int x); 
int n,i; 
Printf (" please enter a positive integer greater than or equal to 3: ", n); 
While (the scanf (" % d ", & amp; N) & amp; & N & lt; 3) {
Printf (" please enter a positive integer greater than or equal to 3! \n"); 
} 
I=prime (n); 
If (I==1) {
Printf (" % d is a prime \ n ", n); 
} else {
Printf (" % d is not a prime \ n ", n); 
} 

return 0; 
} 

Int prime (int x) {
int y; 
If (x & lt;=3) return x & gt; 1; 
For (y=2; Y & lt; X - 1; + + y) 
If (x % y==0) 
return 0; 
return 1; 
} 

CodePudding user response:

Fixed line 8 upstairs, much n

 printf (" please enter a positive integer greater than or equal to 3: "); 

For 1 * 2 * 3 * 4 * 5, you can also write so:

//o 1 * 2 * 3 * 4 * 5: 

#include 
Int main () {
Int t=1;//multiplier 
Int I=5;//the multiplicand 
While (t *=I, I -, i> 0); 
Printf (" % d ", t); 
return 0; 
}