2016-10-27

1. Write the program, enter an integer x, and follow the output to the corresponding Y-value.

 

#include <stdio.h>intMain () {intI,x,y; printf ("Please enter an integer x\n"); scanf ("%d",&x); if(%2!=0) {y=0;  for(i=1; i<=x;i+=2) {y=y+i; } printf ("%d\n", y); }        Else if(%2==0) {y=0;  for(i=2; i<=x;i+=2) {y=y+i; } printf ("%d\n", y); }        return 0;}

Summary: Attention to the discussion of the situation2. Programming 1 -1/2+1/3 -1/4+1/5 - ... +1/99 -1/100 , the result is two decimal places. 　

#include <stdio.h>intMain () {inti; floatsum; Sum=0;  for(i=1; i<= -; i++)    {        if(i%2!=0) {sum=sum-(1.0/i); }        Else if(i%2==0) {sum=sum+ (1.0/i); }} printf ("and for%.2f\n", sum); return 0;}

Summary: Be aware that negative sums are reduced

3. output The era name of all leap years from year to year , and each output is a line of era name. Finally count the total number of leap years.

#include <stdio.h>intMain () {inta,b=0;  for(a= -; a<= the; a++)   {       if(a%4==0&&a% -!=0|| a% -==0) {printf ("%d", a); b++; if(b%Ten==0) {printf ("\ n"); }}} printf ("There are a total of%d leap years \ n", b);return 0;}

4. Enter a real number x and an integer mto calculate XM and do not allow the POW() function to be called.

#include <stdio.h>int  main () {   int  a,b,c,sum;   scanf ("%d%d",&a,&b);     for (sum=1, c=1; c<=b;c++)   {       sum=sum*A;   }   printf ("%d\n", sum);    return 0 ;}

5. enter a string of characters, respectively, to count the number of letters, spaces, numbers, and other characters.

#include <stdio.h>intMain () {intA=0, b=0, c=0, d=0; Charch1; printf ("Please enter a string of characters \ n");  while((Ch1=getchar ())! ='\ n')    {        if((ch1>='A'&&ch1<='Z')|| ch1>='a'&&ch1<='Z') {a++; }        Else if(ch1==' ') {b++; }        Else if(ch1>='0'&&ch1<='9') {C++; }        Else{d++; }} printf ("letters, spaces, numbers, and other characters are%d,%d,%d,%d\n, respectively .", a,b,c,d); return 0;}

Summary: Ch1=getchar () in the relational operator! = The precedence is before the assignment operator = Priority High pressure heater () before the preceding

6. Enter a batch number (positive and negative), enter 0 to end, calculate the average of the positive and negative values, respectively ,

#include <stdio.h>intMain () {inta,b=1, c=1, sum1,sum2; floatAverage1,average2; printf ("Please enter a batch number \ n"); scanf ("%d",&a); Sum1=0; Sum2=0;  while(a!=0)    {        if(a>0) {sum1+=A; b++; }        Else{sum2+=A; C++; } scanf ("%d",&a); }        if(B-1==0) {printf ("no positive number \ n"); }        Else{average1=(float) sum1/(b-1); printf ("positive average is%.2f\n", Average1); }            if(C-1==0) {printf ("no negative number \ n"); }        Else{average2=(float) sum2/(C-1); printf ("negative mean is%.2f\n", Average2); }        return 0;}

Summary: Note that sum must be assigned an initial value

7. output all primes within the range, each line of ten , and the final output of a total number of primes. (aligned per column)

#include <stdio.h>intMain () {intA,b,c=0;  for(a=2; a<= +; a++)    {         for(b=2; b<=a-1; b++)            if(a%b==0)                 Break; if(a==b) {printf ("%03d", B); C++; if(c%Ten==0) {printf ("\ n"); }}} printf ("There's a total of%d prime numbers \ n", c); return 0;}

8. Print the following graphic

             

#include <stdio.h>intMain () {inti,j,k;  for(i=0; i<=4; i++)    {         for(j=0; j<=i;j++) printf (" ");  for(k=1; k<=9-2*i;k++) printf ("*"); printf ("\ n"); }    return 0;}

Summary: Find out what is the first output of the law

Knowledge Point Summary: 1 Note that the cycle must have an initial value

The 2①while statement executes the loop body after the expression is first judged, and the loop body may not execute at once

The ②do while statement executes the loop body and then evaluates the expression, performing at least one loop body

3 The while after the Do and has;

4 Note the precedence of the operator to first perform a low-priority addition ()

2016-10-27