catalogue
Concept blank
5
code
#include<stdio.h>
int main()
{
int k, x;
for (k = 0, x = 0; k <= 9 && x != 10; k++) {
x = x + 2;
}
printf("%d,%d", k, x);
return 0;
}Text description method
1. Define the integer k,x;
2.k=0,x=0
3. Judge K < = 9&&x= 10. If K < = 9 & &x= 10. Execute 3, 4 and 5, otherwise turn to 6;
4.x=x+2
5.k + + to 3
5. Output the value of k,x
flow chart
Operation results
6
code
#include<stdio.h>
int main()
{
char c;
for (c = getchar(); getchar() != '#'; c = getchar())
putchar(c);
return 0;
}Text description method
1. Define character c
2.c = getchar()
3. Judge getchar()! = '#', If getchar()! = '#', Execute 3,4,5, otherwise turn to 6
4.putchar(c)
5.c = getchar()
6. Output c
flow chart
Operation results
Single choice questions
14
code
#include<stdio.h>
int main()
{
char ch;
for (; (ch = getchar()) != '\n';) {
printf("%c", ch);
}
return 0;
}Text description method
1. Define the character ch
2. Judge ch = getchar())! = '\ N ', if ch = getchar())! ='\ N ', execute 2, 3, otherwise turn to 3
3. Output ch
flow chart
Operation results
15
code
#include<stdio.h>
int main()
{
int i, a = 1;
unsigned j;
for (i=15;i>0;i-=2)
a++;
printf("output a=%d", a);
return 0;
}Text description method
1. Define int i, a = 1;unsigned j;
2.i=15
3. Judge I > 0. If I < 0, execute 3 and 4, otherwise turn to 5
4.a + + to 3
5. Output a
flow chart
Operation results
Reading program questions
1
code
#include<stdio.h>
int main()
{
int i;
for (i = 1; i < 6; i++) {
if (i % 2) {
printf("*");
}
else {
printf("#");
}
}
return 0;
}Text description method
1. Define int i
2.i=1
3 judge I < 6, if I < 6, execute 3, 4 and 5, otherwise turn to 6
4. If i%2, output*
5. Otherwise output#
6. Conclusion
flow chart
Operation results
2
code
#include<stdio.h>
int main()
{
int m = 1, n, i;
for (i = 0; i < 5; i++) {
int m = 1;
m++;
if (i == 4) {
n = m;
}
}
printf("%d,%d", m, n);
return 0;
}Text description method
1. Definition: m=1,n,i
2.i=0
3. Judge I < 5. If I < 5, execute 3,4,5,6, otherwise turn to 7
4. Integer m=1
5.m++
6. If i=4, n=m
7. Output m, n
flow chart
Operation results
3
code
#include<stdio.h>
int main()
{
int i;
for (i = 'a'; i < 'f'; i++, i++) {
printf("%c", i - 'a' + 'A');
}
return 0;
}Text description method
1. Define integer i
2.i='a'
3. Judge I <'f '. If I <'f', execute 3, 4 and 5, otherwise turn to 6
4. Output the value of I -'a '+'a'
5.i++,i++
6. Conclusion
flow chart
Operation results
4
code
#include<stdio.h>
int main()
{
int f, f1, f2, i;
f1 = 1; f2 = 1;
printf("%2d %2d", f1, f2);
for (i = 3; i <= 5; i++) {
f = f1 + f2;
printf("%2d", f);
f1 = f2;
f2 = f;
}
return 0;
}Text description method
1. Define the integers f,f1,f2,i
2.f1=1,f2=1
3. Enter the values of F1 and F2
4.i=3
5. Judge I < = 5, if I < = 5, execute 5, 6, 7, 8, 9, otherwise turn to 10
6.f=f1+f2
7. Enter the value of f
8.f1=f2;f2=f
9.i++
10. End
flow chart
Operation results
5
code
#include<stdio.h>
int main()
{
int n = 5;
do
{
switch (n % 2) {
case 0:
n--;
break;
case 1:
n--;
continue;
}
n--;
printf("%2d", n);
} while (n > 0);
return 0;
}Text description method
1. Define integer n=5
2. When n%2
3.case0: n --, turn 7
4.case1:n -- execute 2, 3, 5, 6, 7
5.n--
6. Output the value of n
7. Judge n > 0. If n > 0, execute 2, 3, 4, 5, 6 and 7, otherwise turn to 8
8. Conclusion
flow chart
Operation results
7
code
#include<stdio.h>
int main()
{
int x = 10;
while (x--);
printf("x=%d\n", x);
return 0;
}Text description method
1. Define integer x=10
2. Calculate x--
3. Output the value of x
flow chart
Operation results
9
code
#include<stdio.h>
int main()
{
int m = 9;
for (; m > 0; m--) {
if (m % 3 == 0) {
printf("%d", --m);
}
}
return 0;
}Text description method
1. Define integer m=9
2. Judge m > 0. If M < 0, execute 2, 3 and 4, otherwise turn to 5
3. If m%3==0
4. Output the value of -- m, otherwise turn to 5
5. End
flow chart
Operation results
10
code
#include<stdio.h>
int main()
{
int x = 8;
for (; x > 0; x--) {
if (x % 3 == 0) {
printf("%d,", x--);
continue;
}
printf("%d", --x);
}
return 0;
}Text description method
1. Define integer x=8
2. Judge x > 0. If x < 0, execute 2, 3 and 4, otherwise turn to 5
3. If x%3==0
4. Output the value of x -- to 2, 3 and 5
5. Output the value of --x
flow chart
Operation results
11
code
#include<stdio.h>
int main()
{
int x = 3;
do {
printf("%3d", x = x - 3);
} while (!x);
return 0;
}Text description method
1. Define integer x=3
2. Output x=x-3
3. Judgment! x. If! X execute 2, 3, otherwise turn 4
4. End
flow chart
Operation results
12
code
#include<stdio.h>
int main()
{
int x = 2;
do {
printf("%3d", !x - 2);
} while (--x);
return 0;
}Text description method
1. Define integer x=2
2. Output! x-2
3. Judge -- x, if -- x executes 2, 3, otherwise turn to 4
4. End
flow chart
Operation results
13
code
#include<stdio.h>
int main()
{
int n = 12345, d;
while (n != 0) {
d = n % 10;
printf("%d", d);
n = n / 10;
}
return 0;
}Text description method
1. Define integer n=12345,d
2. When n= 0
3.d=n%10
4. Value of output d
5.n=n/10
6. Conclusion
flow chart
Operation results
14
code
#include<stdio.h>
int main()
{
int m = 0, sum = 0;
char c, oldc = '+';
do {
c = getchar();
if (c <= '9' && c >= '0') {
m = 10 * m + c - '0';
}
else {
if (oldc == '+') {
sum += m;
}
else {
sum -= m;
}
m = 0;
oldc = c;
printf("%3d", sum);
}
} while (c != '=');
return 0;
}Text description method
1. Define integer m=0,sum=0, character c,oldc = '+'
2.c=getchar()
3. If C < ='9 '& & C > ='0', execute m=10*m+c-'0 '
4. Otherwise, if oldc = = '+', execute sum+=m
5. Otherwise sum-=m
6.m=0
7.oldc=c
8. Output the value of sum
9. Judgment c! = '=' Execute 2, 3, 4, 5, 6, 7, 8 and 9, otherwise turn to 10
10. End
flow chart
Operation results
16
code
#include<stdio.h>
int main()
{
int t = 1, n = 235;
do {
t *= n % 10;
n /= 10;
} while (n);
printf("%d\n", t);
return 0;
}Text description method
1. Define integer t=1,n=235
2.t*=n%10
3.n/=10
4. Judge n, execute 2, 3, 4, otherwise turn to 5
5. Value of output t
6. Conclusion
flow chart
Operation results
17
code
#include<stdio.h>
int main()
{
int m = 5, n = 0;
while (m > 0) {
switch (m) {
case 1:
case 3:n += 1; m--; break;
default:n = 0; m--;
case 2:
case 4:n += 2; m--; break;
}
printf("%2d", n);
}
return 0;
}Text description method
1. Integer m=5,n=0
2. When m > 0
3. Judgment m
4. Execute case1:
5. Execute case3:
n+=1,m --, if n+=1 turns 9
6. Default n=0, m--
7. Execute case2:
8. Execute case4:
n+=2, m --, if n+=2 turns 9
9. Output the value of n
10. End
flow chart
Operation results
18
code
#include<stdio.h>
int main()
{
int i, m = 0;
for (i = 0; i < 5; i++) {
switch (i) {
case 0:
case 1:m++;
case 3:m++;
case 4:m--; break;
}
}
printf("%d\n", m);
return 0;
}Text description method
1. Define integer i,m=0
2.i=0
3. Judge i, otherwise turn 10
4. Execute case0:
5. Execute case1:
m++
6. Execute case3:
m++
7. Execute case4: m -- if M -- turn to 8
8. Output the value of m
9.i + + to 3
10. End
flow chart
Operation results
19
code
#include<stdio.h>
int main()
{
int i, b = 0, c = 2;
for (i = 0; i < 2; i++) {
switch (++b, b * c) {
case 1:printf("1");
case 2:printf("2");
case 3:printf("3"); break;
default:printf("other\n");
}
}
return 0;
}Text description method
1. Define integer b=0,c=2
2.i=0
3. Judge I < 2, otherwise turn to 10
4. When + + B, b*c
5. Execute case1:
Output 1
6. Execute case2:
Output 2
7. Execute case3:
Output 3
8. Default output other
9.i + + to 3
10. End
flow chart
Operation results
21
code
#include<stdio.h>
int main()
{
char c;
while ((c = getchar()) != '?')
putchar(--c);
return 0;
}
Text description method
1. Define character c
2. When c = getchar())! = '?'
3.putchar(--c);
4. End
flow chart
Operation results
Perfect program questions
1
code
#include<stdio.h>
int main()
{
int denominator, flag, i, n;
double item, sum;
printf("Please enter n Value of:");
scanf_s("%d", &n);
flag = 1;
denominator = 1;
sum = 0;
for (i = 1; i <= n; i++) {
item = flag * 1.0 / denominator;
sum = sum + item;
flag = -flag;
denominator += 2;
}
printf("sum=%f\n", sum);
return 0;
}
Text description method
1. Define integer denominator, flag, i, n, decimal item, sum
2. Enter the value of n
3.flag = 1
4.denominator = 1
5.sum = 0
6.i=1
7. Judge I < = n, execute 7, 8, 9, 10, 11, 12, otherwise turn to 13
8.item = flag * 1.0 / denominator;
9.sum = sum + item
10.flag = -flag
11.denominator += 2
12.i++
13. Output sum value
14. Conclusion
flow chart
Operation results
2
code
#include<stdio.h>
int main()
{1
int i, n;
float sum = 0, flag = 1;
scanf_s("%d", &n);
for (i = 1; i <= n; i++) {
sum = sum + (flag * i) / (2 * i - 1);
flag = -flag;
}
printf("Sum=%f\n", sum);
return 0;
}Text description method
1. Define integer i,n, decimal sum = 0, flag = 1
2. Output the value of n
3.i=1
4. Judge I < = n, if I < = n, execute 4, 5, 6, 7, otherwise turn to 8
5.sum = sum + (flag * i) / (2 * i - 1)
6.flag = -flag
7.i++
8. Output the value of sum
9. Conclusion
flow chart
Operation results
3
code
#include<stdio.h>
int main()
{
int n, s;
printf("Please enter a number:");
scanf_s("%d", &n);
printf("Output:");
do {
s = n % 10;
printf("%d", s);
n /= 10;
} while (n != 0);
return 0;
}
Text description method
1. Define the integer n,s
2. Please enter a number for n
3. Output
4.s = n % 10;
5.printf("%d", s);
6.n /= 10;
7. Judge n= 0 if n= 0 execute 4, 5, 6, 7, otherwise turn to 8
8. Conclusion
flow chart
Operation results
4
code
#include<stdio.h>
int main()
{
int k,n;
double s;
s = 1.0;
k = 1;
scanf_s("%d", &n);
while (k <= n) {
s = s + 1.0 / (k * (k + 1));
k++;
}
printf("s=%f\n\n", s);
return 0;
}Text description method
1. Define the integer k,n; Decimal s
2.s = 1.0;
3.k = 1;
4. Enter the value of n
5. Judge K < = n, if K < = n, execute 5, 6, 7, otherwise turn to 8
6.s = s + 1.0 / (k * (k + 1))
7.k++
8. Output the value of s
9. Conclusion
flow chart
Operation results
10
code
#include<stdio.h>
int main()
{
int i;
for (i = 100; i < 200; i++) {
if ((i - 2) % 4 == 0) {
if (!((i - 3) % 7)) {
if ((i - 5) % 9 == 0) {
printf("%d", i);
}
}
}
}
return 0;
}
Text description method
1. Define integer i
2.i=100
3. Judge I < 200. If I < 200, execute 3, 4, 5, 6, 7 and 8, otherwise turn to 9
4. If (i - 2)% 4 = 0), execute 5, otherwise i + + to 3
5. If! ((i - 3)% 7 execute 6, otherwise i + + to 3
6. If (i - 5)% 9 = = 0, execute 7, otherwise i + + to 3
7. Value of output i
8.i + + to 3
9. Conclusion
flow chart
Operation results
11
code
#include<stdio.h>
int main()
{
int count, i, n;
double grade, total;
printf("Enter n:");
scanf_s("%d", &n);
total = 0;
count = 0;
for (i = 1; i <= n; i++) {
printf("Enter grade #%d:", i);
scanf_s("%lf", &grade);
total = total + grade;
if (grade < 60) {
count++;
}
}
printf("Grade average=%.2f\n", total / n);
printf("Number of failures=%d\n", count);
return 0;
}Text description method
1. Define integer count, i, n, decimal grade, tota
2. Enter a value for Enter n
3.total = 0
4.count = 0
5.i=1
6. Judge I < = n, if I < = n, execute 6, 7, 8, 9, 10, otherwise turn to 11
7. Output the value of grade
8.total = total + grade
9. If grade < 60, execute count++
10.i + + to 6
11. Output the value of total
12. Output the value of count
13. Conclusion
flow chart
Operation results
12
code
#include<stdio.h>
int main()
{
int i, j;
for (i = 0; i < 10; i++) {
j = i * 10 + 6;
if (j % 3 == 0) {
printf("%3d", j);
}
}
return 0;
}Text description method
1. Define the integer i,j
2.i=0
3. Judge if I < 10, execute if I < 10, otherwise turn to
4.j = i * 10 + 6
5. If J% 3 = 0, execute 6
6. Value of output j
7.i + + to 3
8. Conclusion
flow chart
Operation results
