c语言期末考试题

1、#include <>main(){ int a, b, square; scanf (“%d%d”, &a, &b); square = a * a + b * b; if (square > 100) \*判断a、b的平方和是否大于100 *、 { printf (“their square is bigger than 100\n”); printf (“the digitale bigger than 100 is :%d”, square / 100); } else { printf (“their square is smaller than 100\n”); printf (“their addtion is: %d”, a + b); }}2、#include <>main(){ int n; if ((n % 5 == 0) && (n % 7 == 0)) \* 判断n是否为5和7的公倍数 *\ { printf (“5 and 7 yes”); } else { if (n % 3 == 0) \* 判断是否能被3整除 *\ printf (“ 3 yes”); else printf (“no”); }}3、(3题和平共处题与第2题相似的，只要把条件改一下就可以了）#include <>main(){ int n; if ((n % 3 == 0) && (n % 5 == 0)) { printf (“3 and 5 yes”); } else { if (n % 7 == 0) printf (“ 7 yes”); else printf (“no”); }}4、#include <>main(){ int n; if ((n % 2 == 0) && (n % 3 == 0)) { printf (“2&3 yes”); } else { if (n % 7 == 0) printf (“ 3 yes”); else printf (“no”); }}5、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x == 2) \* 用一个多分支语句将几种情况分开计算*\ y = 2 * x; else if (x < 2) y = x * x + 1; else y = 2 * x * x + 3 * x + 1; printf ("y = %d", y);}6、（6 题和7题还有8题都与5题相似）#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x == 1) y = 1; else if (x < 1) y = x * x; else y = x * x * x; printf ("y = %d", y);}7、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x < 0) y = x + 1; else if (x <= 2) y = x * x + 2; else y = x * x * x + 3; printf ("y = %d", y);}8、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x < 1) y = x; else if (x <= 10) y =2 * x - 1; else y = 3 * x - 11; printf ("y = %d", y);}9、#include <>main(){ float score; char grade; printf ("please enter the score:"); scanf ("%f", &score); if (score > 100 || score < 0) \*判断成绩是否输入正确*\ { printf ("enter error!"); } if (score >= 90) grade = 'A'; else if (score >= 80) \*在这里的else if 语句条件中已经排除了在于等于90的情况，即此处等同于"score >= 80 && score < 90"*\ grade = 'B'; else if (score >= 70) grade = 'C'; else if (score >= 60) grade = 'D'; else grade = 'E'; printf ("the grade is :%c", grade);}10、#include <>main(){ int n, price; printf ("please enter quantity:"); scanf ("%d", &n); if (n <= 10) price = 60; else if (n < 40) \* 此处方法与上题相同*\ price = 50; else price = 45; printf ("the total money is :%d", price * n);}11、#include <>#include <>main(){ int a, b,final; printf ("enter a,b:"); scanf ("%d%d", &a, &b); if (a % b == 0) final = a * a + b * b; else if ( b % a == 0) final = a * a * a + b * b * b; else final = abs(a - b); printf ("final = %d", final);}12、#include <>#include <>main(){ float a, b, f; printf ("enter a,b:"); scanf ("%f%f", &a, &b); if (a > b) f = fabs(a - b); else if (a < b) f = a * b; else f = ((int)a % 10) * ((int)b % 10); \*用强制转换将a、b转换成整数再除10求余即得个位数字*\ printf ("f = %.2f", f); }13、 #include <>#include <>main(){ int a, b, c, disc; int x1, x2，x; printf ("a,b,c="); scanf ("%d%d%d", &a, &b, &c); if (a == 0) { printf ("this is not a equation"); } else { disc = b * b - 4 * a * c; if ( disc > 0) { x1 = （- b + sqrt(disc)) / （2 * a); x2 = - b - sqrt(disc) / (2 * a); printf ("there are two deferent root:%d %d", x1, x2）； } else { x = - b / ( 2 * a); printf ("there are two same root:%d", x); } }}14、#include <>main(){ int n; int n1, n2, n3; printf ("please enter a number:"); scanf ("%d", &n); if (n >= 100 && n <= 999) { n1 = n / 10;\*取出n的个、十、百各位数*\ n2 = (n - n1* 10) / 10； n3 = n % 10; if (n1 * n1 * n1 + n2 * n2 * n2 + n3 * n3 * n3 == n)\* 判断个、十、百位平方和是否等于n *\ printf ("the number is shuixianhua digitale"); else printf ("the number is not shuixianhua digitale"); } else printf ("the number is not a 3 bit number");}15、#include <>main(){ enum workday {monday = 1, tuesday, wednesday, thursday, friday, saturday, sunday}; enum workday workdays; int n; printf ("please enter a days:"); scanf ("%1d", &n); if (n >= 1 && n <= 7) { workdays = (enum workday)n; switch(workdays) { case 1: printf ("moday"); break; case 2: printf ("tuesday"); break; case 3: printf ("wednesday"); break; case 4: printf ("thursday"); break; case 5: printf ("friday"); break; case 6: printf ("saturday"); break; case 7: printf ("sunday"); break; } } else printf ("enter error");} 16、#include <>main(){ int i, n; float t, s = 0; \* 用t 产生各项，s 为各项之和 *\ printf ("n = "); scanf ("%d", &n); for (i = 1; i <= n; i++) { t = / n; \* 因为n 是整数，在1后面加小数使得结果不至于为零*\ s = s + t; n = n + 2; } printf ("the addtion is :%f", s);}

这么简单的程序自己怎么不写啊，这是多么基础的啊！

#include <>

#include <>

#include <>

int main()

{

int n,*a,i,j,t;

srand(time(NULL));

do

{

scanf("%d",&n);

}while(n<1);

a=(int*)malloc(sizeof(int)*n);

for(i=0;i

a[i]=rand()%99+1;

for(i=0;i

printf("%d%c",a[i],(i+1)%10?' ':'\n');

printf("\n");

for(i=0,j=n-1;i

{

t=a[i];

a[i]=a[j];

a[j]=t;

}

for(i=0;i

printf("%d%c",a[i],(i+1)%10?' ':'\n');

printf("\n");

free(a);

return 0;

}

1输入两个整数a和b，若a和b的平方和大于100，则输出平方和的百位以上的数字，否则输出a和 b的和。 要求： 1）输出结果时说明平方和是大于100还是小于100（ >100或<100 ）#include<>int main(）{ int a, b, c; scanf("%d%d", &a, &b); c = a * a + b * b; if(c > 100) printf("平方和大于100\n%d\n", c%100); else printf("平方和小于等于100\n%d\n", a + b); return 0;} 2输入一个整数，判断是否是5和7的公倍数，若是则输出：5and7yes，否则再判断是否是3的倍数，若是3的倍数输出：3yes，若都不是则输出：no#include<>int main(){ int input; scanf("%d", &input); if(input % 5 == 0 && input % 7 == 0)printf("5and7yes\n"); else if(input % 3 == 0) printf("3yes\n"); else printf("no\n"); return 0;}第三，四题和第二题差不多 5计算公式： [ 2*x x=2 y= [ x*x+1 x<2 [ 2*x*x+3*x+1 x>2 要求： 1）从键盘输入x的值，根据x的值求y的值 2）输出y的值#include<>int main(){ int x, y; scanf("%d", &x); if(x == 2) y = 2*x; else if(x < 2) y = x*x+1; else y = 2 * x * x + 3 * x + 1; printf("%d\n", y); return 0;}8 和7差不多 下面原理都差不多 用if基本都能搞定 判断条件就行了 应该LZ能搞定了 东西有点多 -。- 就写这些吧 其他的就不一一写了^ ^