Nanjing Normal University, a re-examination of the training program

1. Enter a, b, c of three values, wherein the maximum of the output

#include <stdio.h>

int f(int a,int b,int c)
{
    int m;
    if(a>b)
        m=a;
    else
        m=b;
    if(c>m)
        m=c;
    return m;
}
int main()
{
    int a,b,c;
    scanf("%d %d %d",&a,&b,&c);
    printf("%d\n",f(a,b,c));
    return 0;
}

operation result:

2. seek 5!

#include <stdio.h>

int main()
{
    int i,t=1;
    for(i=2; i<=5; i++)
        t=t*i;
    printf("5!=%d",t);
    return 0;
}

operation result:

3 + ... find polynomial 1-1 / 2 + 1 / 3-1 / 4 + 1 / 99-1 / 100

#include <stdio.h>

int main()
{
    int i,sign=1;
    double sum=1.0,term;
    for(i=2; i<=100; i++)
    {
        sign=-sign;
        term=1.0/i;
        sum+=term;
    }
    printf("%lf\n",sum);
    return 0;
}

operation result:

4. The area of ​​the triangle seek

#include <stdio.h>
#include <math.h>
int main()
{
    double a,b,c,s,area;
    scanf("%lf %lf %lf",&a,&b,&c);
    s=(a+b+c)/2;
    area=sqrt(s*(s-a)*(s-b)*(s-c));
    printf("a=%.2lf\tb=%.2lf\tc=%.2lf\n",a,b,c);
    printf("area=%.2lf\n",area);
    return 0;
}

operation result:

The root seeking a * x² + b * x + c = 0 Equation. a, b, c by a keyboard input, provided b²-4 * a * c> 0

#include <stdio.h>
#include <math.h>
int main()
{
    double a,b,c,disc,x1,x2,p,q;
    scanf("%lf %lf %lf",&a,&b,&c);
    disc=b*b-4*a*c;
    p=-b/(2.0*a);
    q=sqrt(disc)/(2.0*a);
    x1=p+q;
    x2=p-q;
    printf("x1=%.2lf\nx2=%.2lf\n",x1,x2);
    return 0;
}

operation result:

6. Use π / 4≈1-1 / 3 + 1 / 5-1 / 7 + ... to approximate formula of [pi], until it finds an absolute value of less than one up to six degrees is 10 (without the accumulation )

#include <stdio.h>
#include <math.h>
int main()
{
    int sign=1;
    double pi=0.0,n=1.0,term=1.0;
    while(fabs(term)>=1e-6)
    {
        pi+=term;
        n=n+2;
        sign=-sign;
        term=sign/n;
    }
    pi=pi*4;
    printf("pi≈%.8lf\n",pi);
    return 0;
}

operation result:

7. 40 before seek Fibinacci number sequence. (One pair of rabbits from the first 3 months after birth are born every month one pair of rabbits, bunnies grow up to the third month gave birth to a pair of rabbits do not assume that all the dead, asked the total number of rabbits per month how many?)

#include <stdio.h>

int main()
{
    int fi[20]= {1,1};
    for(int i=2; i<20; i++)
        fi[i]=fi[i-1]+fi[i-2];
    for(int i=0; i<20; i++)
    {
        if(i%5==0)
            printf("\n");
        printf("%12d",fi[i]);
    }
    printf("\n");
    return 0;
}

operation result:

8. Enter a number, it is determined whether or not a prime number

#include <stdio.h>
#include <math.h>
int main()
{
    int n,flag=1;
    scanf("%d",&n);
    for(int i=2; i<=sqrt(n); i++)
    {
        if(n%i==0)
        {
            flag=0;
            break;
        }
    }
    if(flag)
        printf("%d is prime!\n",n);
    else
        printf("%d is not prime!\n",n);
    return 0;
}

operation result:

9. Finding all prime numbers between 100 and 200

#include <stdio.h>
#include <math.h>
int isPrime(int n)  //判断是否为素数
{
    for(int i=2; i<=sqrt(n); i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}
int main()
{
    for(int i=100; i<=200; i++)
    {
        if(isPrime(i))
            printf("%d\n",i);
    }

    return 0;
}

operation result:

10. Translation password A-> E, a-> e, i.e. the subsequent letters are converted to four letters

W->A，X->B，Y->C，Z->D，

Non-alphabetic maintain the status quo unchanged

For example: "China" -> "Glmre"

#include <stdio.h>

int main()
{
    char c;
    while((c=getchar())!='\n')
    {
        if((c>='A'&&c<='Z')||(c>='a'&&c<='z'))
        {
            c=c+4;
            if((c>'Z'&&c<'Z'+4)||c>'z')
                c=c-26;
        }
        printf("%c",c);
    }
    printf("\n");
    return 0;
}

operation result: