1、浙江大学C程试题库更新c语言程序设计题目及答案 20021程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:输入 x,计算并输出下列分段函数 f(x) 的值(保留1位小数)。当 x 不等于0时,y = f(x) = 1/x,当 x 等于0时,y = f(x) = 0。输入输出示例:括号内是说明输入2 (repeat=2)10 (x=10)0 (x=0)输出f(10.00) = 0.1f(0.00) = 0.0#include int main(void) int repeat, ri; double x, y; scanf(%
2、d, &repeat); for(ri = 1; ri = repeat; ri+) scanf(%lf, &x); /*-*/ if(x!=0) y=1/x;else y=0; printf(f(%.2f) = %.1fn, x, y); 20022程序填空,不要改变与输入输出有关的语句。输入华氏温度,输出对应的摄氏温度。计算公式:c = 5*(f-32)/9,式中:c表示摄氏温度,f表示华氏温度。输入输出示例:括号内为说明输入150 (fahr=150)输出celsius = 65#include int main(void)int celsius, fahr; /*-*/scanf(“%
3、d”,&fahr); celsius=5.0*(fahr-32)/9; printf(celsius = %dn, celsius); 20023程序填空,不要改变与输入输出有关的语句。输入存款金额 money、存期 year 和年利率 rate,根据下列公式计算存款到期时的利息 interest(税前),输出时保留2位小数。interest = money(1+rate)year - money输入输出示例:括号内为说明输入1000 3 0.025 (money = 1000, year = 3, rate = 0.025)输出interest = 76.89#include #includ
4、e int main(void) int money, year;double interest, rate;/*-*/scanf(“%d%d%lf”,&money,&year,&rate); interest=money*pow(1+rate),year)-money; printf(interest = %.2fn, interest); 20024程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat= 0时,f(x) = x0.5,当x小于0时,f(x) = (x+1)2 + 2x + 1/x。输入输出示例:括号内是说明输入3 (repeat=3)10-0
5、.50输出f(10.00) = 3.16f(-0.50) = -2.75f(0.00) = 0.00#include #include int main(void) int repeat, ri; double x, y; scanf(%d, &repeat); for(ri = 1; ri =0) y=sqrt(x);else y=pow(x+1),2)+2*x+1/x; printf(f(%.2f) = %.2fn, x, y); 20025程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:输入实数 x,计算并输出下列分段
6、函数 f(x) 的值,输出时保留1位小数。当 x 不等于10时,y = f(x) = x,当 x 等于10时,y = f(x) = 1/x。输入输出示例:括号内是说明输入2 (repeat=2)10234输出f(10.0) = 0.1f(234.0) = 234.0#include int main(void) int repeat, ri; double x, y; scanf(%d, &repeat);for(ri = 1; ri = repeat; ri+) /*-*/scanf(%lf, &x);if(x!=10) y=x;else y=1/x; printf(f(%.1f) = %.
7、1fn, x, y); 20026程序填空,不要改变与输入输出有关的语句。输入2个整数 num1 和 num2,计算并输出它们的和、差、积、商与余数。输出两个整数的余数可以用 printf(%d % %d = %dn, num1, num2, num1%num2);输入输出示例:括号内是说明输入5 3 (num1=5,num2=3)输出5 + 3 = 85 - 3 = 25 * 3 = 155 / 3 = 15 % 3 = 2#include int main(void)int num1, num2; /*-*/scanf(%d%d, &num1,&num2); printf(%d + %d
8、= %dn, num1, num2, num1+num2); printf(%d - %d = %dn, num1, num2, num1-num2); printf(%d * %d = %dn, num1, num2, num1*num2); printf(%d / %d = %dn, num1, num2, num1/num2); printf(%d % %d = %dn, num1, num2, num1%num2); return 0;20031程序填空,不要改变与输入输出有关的语句。计算表达式 1 + 2 + 3 + . + 100的值。输出示例:sum = 5050 #includ
9、e int main(void) int i, sum;/*-*/sum=0;for(i=1;i=100;i+) sum=sum+i; printf(sum = %dn, sum); 20032程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:输入一个正整数m(0=m=100),计算表达式 m + (m+1) + (m+2) + . + 100的值。输入输出示例:括号内为说明输入3 (repeat=3)0 (计算0+1+2+.+100)10 (计算10+11+12+.+100)50 (计算50+51+52+.+100)输出su
10、m = 5050sum = 5005sum = 3825 #include int main(void) int i, m, sum; int repeat, ri; scanf(%d, &repeat); for(ri = 1; ri = repeat; ri+) scanf(%d, &m); /*-*/ sum=0; for(i=m;i=100;i+) sum=sum+i; printf(sum = %dn, sum); 20033程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:输入2个正整数 m 和 n(m=n),计算
11、表达式 1/m + 1/(m+1) + 1/(m+2) + . + 1/n的值,输出时保留3位小数。输入输出示例:括号内为说明输入3 (repeat=3)5 15 (计算1/5+1/6+1/7+.+1/15)10 20 (计算1/10+1/11+1/12+.+1/20)1 3 (计算1+1/2+1/3)输出sum = 1.235sum = 0.769sum = 1.833 #include int main(void) int i, m, n; int repeat, ri; double sum; scanf(%d, &repeat); for(ri = 1; ri = repeat; ri
12、+) scanf(%d%d, &m, &n); /*-*/ sum=0; for(i=m;i=n;i+) sum=sum+1.0/i; printf(sum = %.3fn, sum); 20034程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:输入一个正整数 n,计算表达式 1 + 1/3 + 1/5 + . 的前 n 项之和,输出时保留6位小数。输入输出示例:括号内为说明输入2 (repeat=2)5 (计算1+1/3+1/5+1/7+1/9)23 (计算1+1/3+1/5+.+1/45)输出sum = 1.787302
13、sum = 2.549541 #include int main(void) int i, n; int repeat, ri; double sum; scanf(%d, &repeat); for(ri = 1; ri = repeat; ri+) scanf(%d, &n);/*-*/ sum=0; for(i=1;i=n;i+) sum=sum+1.0/(2*i-1); printf(sum = %.6fn, sum); 20035程序填空,不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10),做repeat次下列运算:读入一个正整数 n,计算11/41/71/10的前 n 项之和,输出时保留3位小数。输入输出示例:括号内是说明输入2 (repeat=2)3
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