浙江大學C程題庫(更新).doc

_c語言程序設計題目及答案 20021程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入 x，計算并輸出下列分段函數(shù) f(x) 的值(保留1位小數(shù))。當 x 不等于0時，y = f(x) = 1/x，當 x 等于0時，y = f(x) = 0。輸入輸出示例：括號內是說明輸入2 (repeat=2)10 (x=10)0 (x=0)輸出f(10.00) = 0.1f(0.00) = 0.0#include <stdio.h>int main(void) int repeat, ri; double x, y; scanf("%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，不要改變與輸入輸出有關的語句。輸入華氏溫度，輸出對應的攝氏溫度。計算公式：c = 5*(f-32)/9，式中：c表示攝氏溫度，f表示華氏溫度。輸入輸出示例：括號內為說明輸入150 (fahr=150)輸出celsius = 65#include <stdio.h>int main(void)int celsius, fahr; /*-*/scanf(“%d”,&fahr); celsius=5.0*(fahr-32)/9; printf("celsius = %dn", celsius); 20023程序填空，不要改變與輸入輸出有關的語句。輸入存款金額 money、存期 year 和年利率 rate，根據(jù)下列公式計算存款到期時的利息 interest(稅前)，輸出時保留2位小數(shù)。interest = money(1+rate)year - money輸入輸出示例：括號內為說明輸入1000 3 0.025 (money = 1000, year = 3, rate = 0.025)輸出interest = 76.89#include <stdio.h>#include <math.h>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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入 x，計算并輸出下列分段函數(shù) f(x) 的值(保留2位小數(shù))，請調用 sqrt 函數(shù)求平方根，調用 pow 函數(shù)求冪。當x >= 0時，f(x) = x0.5，當x小于0時，f(x) = (x+1)2 + 2x + 1/x。輸入輸出示例：括號內是說明輸入3 (repeat=3)10-0.50輸出f(10.00) = 3.16f(-0.50) = -2.75f(0.00) = 0.00#include <stdio.h>#include <math.h>int main(void) int repeat, ri; double x, y; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+)/*-*/scanf("%lf", &x);if(x>=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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入實數(shù) x，計算并輸出下列分段函數(shù) f(x) 的值，輸出時保留1位小數(shù)。當 x 不等于10時，y = f(x) = x，當 x 等于10時，y = f(x) = 1/x。輸入輸出示例：括號內是說明輸入2 (repeat=2)10234輸出f(10.0) = 0.1f(234.0) = 234.0#include <stdio.h>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) = %.1fn", x, y); 20026程序填空，不要改變與輸入輸出有關的語句。輸入2個整數(shù) num1 和 num2，計算并輸出它們的和、差、積、商與余數(shù)。輸出兩個整數(shù)的余數(shù)可以用 printf("%d % %d = %dn", num1, num2, num1%num2);輸入輸出示例：括號內是說明輸入5 3 (num1=5,num2=3)輸出5 + 3 = 85 - 3 = 25 * 3 = 155 / 3 = 15 % 3 = 2#include <stdio.h>int main(void)int num1, num2; /*-*/scanf("%d%d", &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); printf("%d % %d = %dn", num1, num2, num1%num2); return 0;20031程序填空，不要改變與輸入輸出有關的語句。計算表達式 1 + 2 + 3 + . + 100的值。輸出示例：sum = 5050 #include <stdio.h>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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入一個正整數(shù)m(0<=m<=100)，計算表達式 m + (m+1) + (m+2) + . + 100的值。輸入輸出示例：括號內為說明輸入3 (repeat=3)0 (計算0+1+2+.+100)10 (計算10+11+12+.+100)50 (計算50+51+52+.+100)輸出sum = 5050sum = 5005sum = 3825 #include <stdio.h>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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入2個正整數(shù) m 和 n(m<=n)，計算表達式 1/m + 1/(m+1) + 1/(m+2) + . + 1/n的值，輸出時保留3位小數(shù)。輸入輸出示例：括號內為說明輸入3 (repeat=3)5 15 (計算1/5+1/6+1/7+.+1/15)10 20 (計算1/10+1/11+1/12+.+1/20)1 3 (計算1+1/2+1/3)輸出sum = 1.235sum = 0.769sum = 1.833 #include <stdio.h>int main(void) int i, m, n; int repeat, ri; double sum; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) 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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入一個正整數(shù) n，計算表達式 1 + 1/3 + 1/5 + . 的前 n 項之和，輸出時保留6位小數(shù)。輸入輸出示例：括號內為說明輸入2 (repeat=2)5 (計算1+1/3+1/5+1/7+1/9)23 (計算1+1/3+1/5+.+1/45)輸出sum = 1.787302sum = 2.549541 #include <stdio.h>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，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：讀入一個正整數(shù) n，計算11/41/71/10的前 n 項之和，輸出時保留3位小數(shù)。輸入輸出示例：括號內是說明輸入2 (repeat=2)310輸出sum = 0.893sum = 0.819 #include <stdio.h>int main(void) int flag, i, n, t; int repeat, ri; double item, sum; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d", &n);/*-*/ sum=0;flag=1;item=0;t=1; for(i=1;i<=n;i+) item=flag*1.0/t;sum=sum+item;flag=-flag;t=t+3; printf("sum = %.3fn", sum); 20036程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：讀入2個整數(shù) lower 和 upper，輸出一張華氏攝氏溫度轉換表，華氏溫度的取值范圍是lower, upper，每次增加2F。計算公式：c = 5 * (f - 32) / 9，其中：c表示攝氏溫度，f表示華氏溫度。輸出請使用語句 printf("%3.0f %6.1fn", fahr, celsius);輸入輸出示例：括號內是說明輸入2 (repeat=2)32 35 (lower=32,upper=35)40 30 (lower=40,upper=30)輸出fahr celsius 32 0.0 34 1.1fahr celsius#include <stdio.h>int main(void) int lower, upper; int repeat, ri; double celsius, fahr; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d%d", &lower, &upper); printf("fahr celsiusn"); /*-*/for(fahr=lower;fahr<=upper;fahr=fahr+2) celsius=5 * (fahr- 32) / 9; printf("%3.0f %6.1fn", fahr, celsius); 20037程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入2 個正整數(shù) m 和 n，計算 m!n!。輸入輸出示例：括號內是說明輸入：2 (repeat=2)1 4 (m=1,n=4)3 8 (m=3,n=8)輸出：1! + 4! = 253! + 8! = 40326#include "stdio.h"int main(void) int i, m, n; int repeat, ri; double fm, fn; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d%d", &m, &n);/*-*/ fm=fn=1; for(i=1;i<=m;i+) fm=fm*i; for(i=1;i<=n;i+) fn=fn*i; printf("%d! + %d! = %.0fn", m, n, fm+fn); 20038程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：讀入1 個實數(shù)x和正整數(shù) n(n<=50)，計算并輸出 x 的 n 次冪(保留2位小數(shù))，不允許調用pow函數(shù)求冪。輸入輸出示例：括號內是說明輸入2 (repeat=2)1.5 2 (x=1.5,n=2)2 7 (x=2,n=7)輸出2.25128.00 #include <stdio.h>int main(void) int i, n; int repeat, ri; double mypow, x; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%lf%d", &x, &n); /*-*/ mypow=1; for(i=1;i<=n;i+) mypow = mypow*x; printf("%.2fn", mypow); 20041程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)n，生成一張3的乘方表，輸出30 3n的值，可調用冪函數(shù)計算3的乘方。輸出使用語句 printf("pow(3,%d) = %.0fn", i, mypow);輸入輸出示例：括號內是說明輸入3 (n=3)輸出pow(3,0) = 1pow(3,1) = 3pow(3,2) = 9pow(3,3) = 27#include <stdio.h>#include <math.h>int main(void) int i, n; double mypow; scanf("%d", &n);/*-*/for(i=0;i<=n;i+) mypow=pow(3,i); printf("pow(3,%d) = %.0fn", i, mypow); return 0;20042程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)n，生成一張階乘表，輸出 1! n! 的值，要求定義和調用函數(shù)fact(n)計算 n!，函數(shù)類型為double。輸出使用語句 printf("%d! = %.0fn", i, myfact);輸入輸出示例：括號內是說明輸入3 (n=3)輸出1! = 12! = 23! = 6 #include <stdio.h>int main(void) int i, n; double myfact; double fact(int n); scanf("%d", &n);/*-*/for(i=1;i<=n;i+) myfact=fact(i); printf("%d! = %.0fn", i, myfact); return 0;/*-*/double fact(int n) int i;double f=1;for(i=1;i<=n;i+)f=f*i;return f; 20043程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入2個正整數(shù) m 和 n(m<=n)，計算 n! /(m!* (n-m)!) 。要求定義并調用函數(shù)fact(n)計算n的階乘, 其中 n 的類型是 int，函數(shù)類型是 double。例：括號內是說明輸入：2 (repeat=2)2 7 (m=2, n=7)5 12 (m=5, n=12)輸出：result = 21result = 792#include "stdio.h"double fact(int n);int main(void) int m, n; int repeat, ri; double s; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d%d", &m, &n);/*-*/ s= fact(n)/(fact(m)*fact(n-m); printf("result = %.0fn", s);return 0;/*-*/double fact(int n) int i;double f=1;for(i=1;i<=n;i+)f=f*i;return f; 20044程序填空，不要改變與輸入輸出有關的語句。計算 1000.51010.510000.5的值(保留2位小數(shù))，可調用sqrt函數(shù)計算平方根。輸入輸出示例：括號內是說明輸出sum = 20435.99#include <stdio.h>#include <math.h>int main(void) int i; double sum; /*-*/ sum=0; for(i=100;i<=1000;i+) sum=sum+sqrt(i); printf("sum = %.2fn", sum);30001程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù)repeat (0<repeat<10)，做repeat次下列運算：輸入?yún)?shù)a,b,c，求一元二次方程a*x*xb*xc0的根，結果保留2位小數(shù)。輸出使用以下語句：printf("參數(shù)都為零，方程無意義!n");printf("a和b為0，c不為0，方程不成立n");printf("x = %0.2fn", -c/b);printf("x1 = %0.2fn", (-b+sqrt(d)/(2*a);printf("x2 = %0.2fn", (-b-sqrt(d)/(2*a);printf("x1 = %0.2f+%0.2fin", -b/(2*a), sqrt(-d)/(2*a);printf("x2 = %0.2f-%0.2fin", -b/(2*a), sqrt(-d)/(2*a);輸入輸出示例：括號內為說明輸入：5 (repeat=5)0 0 0 (a=0,b=0,c=0)0 0 1 (a=0,b=0,c=1)0 2 4 (a=0,b=2,c=4)2.1 8.9 3.5 (a=2.1,b=8.9,c=3.5)1 2 3 (a=1,b=2,c=3)輸出：參數(shù)都為零，方程無意義!a和b為0，c不為0，方程不成立x = -2.00x1 = -0.44x2 = -3.80x1 = -1.00+1.41ix2 = -1.00-1.41i#include <stdio.h>#include <math.h>int main(void) int repeat, ri; double a, b, c, d; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%lf%lf%lf", &a, &b, &c);/*-*/ d=b*b-4*a*c;if(a=0)if(b=0)if(c=0) printf("參數(shù)都為零，方程無意義!n"); else printf("a和b為0，c不為0，方程不成立n"); else printf("x = %0.2fn", -c/b); else if(d>=0)printf("x1 = %0.2fn", (-b+sqrt(d)/(2*a);printf("x2 = %0.2fn", (-b-sqrt(d)/(2*a); else printf("x1 = %0.2f+%0.2fin", -b/(2*a), sqrt(-d)/(2*a);printf("x2 = %0.2f-%0.2fin", -b/(2*a), sqrt(-d)/(2*a); 30002程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù) repeat (0<repeat<10)，做 repeat 次下列運算：輸入一個整數(shù) x，計算并輸出下列分段函數(shù) sign(x) 的值。 -1 x < 0y = sign(x) = 0 x = 0 1 x > 0 輸入輸出示例：括號內是說明輸入3 (repeat=3) 10 (x=10) 0 (x=0) -98 (x=-98) 輸出sign(10) = 1 (x = 10時 y = 1)sign(0) = 0 (x = 0時 y = 0)sign(-98) = -1 (x = -98時y = -1)#include <stdio.h>int main(void) int repeat, ri; int x, y; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d", &x); /*-*/ if(x>0) y=1;else if(x=0) y=0;else y=-1; printf("sign(%d) = %dn", x, y);return 0;30003程序填空，不要改變與輸入輸出有關的語句。輸入10個字符，統(tǒng)計其中英文字母、空格或回車、數(shù)字字符和其他字符的個數(shù)。輸入輸出示例：括號內是說明輸入Reold 123?輸出letter = 5, blank = 1, digit = 3, other = 1#include <stdio.h>int main(void) char c; int blank, digit, i, letter, other; blank = digit = letter = other = 0; for(i = 1; i <= 10; i+) c = getchar();/*-*/ if(c >= 'a' && c <= 'z' ) | ( c >= 'A' && c <= 'Z') letter +; else if(c >='0'&&c<='9') digit +; else if(c = ' ' | c = 'n') blank +; else other +; printf("letter = %d, blank = %d, digit = %d, other = %dn", letter, blank, digit, other);return 0;30004程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù) repeat (0<repeat<10)，做 repeat 次下列運算：輸入五級制成績(AE)，輸出相應的百分制成績(0100)區(qū)間，要求使用switch語句。五級制成績對應的百分制成績區(qū)間為：A(90-100)、B(80-89)、C(70-79)、D(60-69)和E(0-59),如果輸入不正確的成績，顯示"Invalid input"。輸出使用以下語句：printf("90-100n"); printf("80-89n"); printf("70-79n"); printf("60-69n");printf("0-59n");printf("Invalid inputn"); 輸入輸出示例：括號內是說明輸入6ABCDEj (repeat=6，輸入的五級成績分別為A、B、C、D、E和無效的字符j) 輸出90-10080-8970-7960-690-59Invalid input (輸入數(shù)據(jù)不合法)#include <stdio.h>int main(void) char ch; int repeat, ri; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) ch = getchar(); /*-*/ switch(ch) case 'A: printf("90-100n");break; case 'B': printf("80-89n");break; case 'C' : printf("70-79n");break;case 'D' : printf("60-69n");break;case 'E' : printf("0-59n"); break; default: printf("Invalid inputn"); break; return 0;30005程序填空，不要改變與輸入輸出有關的語句。查詢水果的單價。有4種水果，蘋果(apples)、梨(pears)、桔子(oranges)和葡萄(grapes)，單價分別是3.00元/公斤，2.50元/公斤，4.10元/公斤和10.20元/公斤。在屏幕上顯示以下菜單(編號和選項)，用戶可以連續(xù)查詢水果的單價，當查詢次數(shù)超過5次時，自動退出查詢；不到5次時，用戶可以選擇退出。當用戶輸入編號14，顯示相應水果的單價(保留1位小數(shù))；輸入0，退出查詢；輸入其他編號，顯示價格為0。輸入輸出示例：括號內是說明輸入3 (oranges的編號) 0 (退出查詢) 輸出1 apples2 pears3 oranges4 grapes0 Exitprice = 4.11 apples2 pears3 oranges4 grapes0 Exit#include <stdio.h>int main(void) int choice, i; double price; for(i = 1; i <= 5; i+) printf("1 applesn"); printf("2 pearsn"); printf("3 orangesn"); printf("4 grapesn"); printf("0 Exitn"); scanf("%d", &choice); if(choice = 0) break; else/*-*/ switch (choice) case 1: price=3.0; break; case 2: price=2.5; break; case 3: price=4.1; break; case 4: price=10.2; break; default: price=0.0; break; printf("price = %0.1fn", price); return 0;30006程序填空，不要改變與輸入輸出有關的語句。輸入5個學生的數(shù)學成績，判斷他們的成績是否及格。如果成績低于60，輸出"Fail"，否則，輸出"Pass"。輸入輸出示例：括號內是說明輸入6159924060輸出PassFailPassFailPass#include <stdio.h>int main(void)0 int i, mark; for(i = 1; i <= 5; i+) scanf("%d", &mark);/*-*/ if(mark>=60) printf(“Passn”); else printf(“Failn”); 30007程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù) repeat (0<repeat<10)，做 repeat 次下列運算：輸入三角形的3條邊 a, b, c，如果能構成一個三角形，輸出面積 area 和周長 perimeter(保留2位小數(shù))；否則，輸出"These sides do not correspond to a valid triangle"。在一個三角形中，任意兩邊之和大于第三邊。三角形面積計算公式：area = (s(s-a)(s-b)(s-c)0.5 ，其中s = (a+b+c)/2輸出使用以下語句：printf("area = %.2f, perimeter = %.2fn",area, perimeter);printf("These sides do not correspond to a valid trianglen"); 輸入輸出示例：括號內是說明輸入2 (repeat=2) 5 5 3 (a=5,b=5,c=3)1 4 1 (a=1,b=4,c=1)輸出area = 7.15, perimeter = 13.00These sides do not correspond to a valid triangle#include <stdio.h>#include <math.h>int main(void) int a, b, c; int repeat, ri; double area, perimeter, s; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%d%d%d", &a, &b, &c);/*-*/ if(a+b>c&&b+c>a&&c+a>b) perimeter=a+b+c; s= perimeter/2;area=sqrt(s*(s-a)*(s-b)*(s-c);printf("area = %.2f, perimeter = %.2fn",area, perimeter); else printf("These sides do not correspond to a valid trianglen");return 0; 30008程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù) repeat (0<repeat<10)，做 repeat 次下列運算：輸入一個職工的月薪 salary，輸出應交的個人所得稅 tax(保留2位小數(shù))。計算公式：tax = rate * (salary - 850)當 salary 850 時，rate = 0當 850 salary 1350 時，rate = 5%當 1350 salary 2850 時，rate = 10%當 2850 salary 5850 時，rate = 15%當 5850 salary 時，rate = 20% 輸入輸出示例：括號內是說明輸入5 (repeat=5)1010.87 (salary=1010.87)32098.76 (salary=32098.76)800 (salary=800)4010 (salary=4010)2850 (salary=2850)輸出tax = 8.04tax = 6249.75tax = 0.00tax = 474.00tax = 200.00#include <stdio.h>int main(void) int repeat, ri; double rate, salary, tax; scanf("%d", &repeat); for(ri = 1; ri <= repeat; ri+) scanf("%lf", &salary);/*-*/ if(salary<=850) rate=0; else if(salary<=1350) rate=0.05; else if(salary<=2850) rate=0.1; else if(salary<=5850) rate=0.15; else rate=0.2; tax = rate * (salary - 850); printf("tax = %0.2fn", tax); 30009程序填空，不要改變與輸入輸出有關的語句。輸入一個正整數(shù) n，再輸入 n 個學生的百分制成績，統(tǒng)計各等級成績的個數(shù)。成績等級分為五級，分別為A(90-100)、B(80-89)、C(70-79)、D(60-69)和E(0-59)。輸入輸出示例：括號內是說明輸入5 (n=5)77 54 92 73 60輸出Number of A(90-100): 1Number of B(80-89): 0Number of C(70-79): 2Number of D(60-69): 1Number of E(0-59): 1#include <stdio.h>int main(void) int i, mark, n; int na, nb, nc, nd, ne; scanf("%d",&n); na = nb = nc = nd = ne = 0; for(i = 1; i <= n; i+) scanf("%d", &mark);/*-*/ if(mark >=90) na+;else if(mark >=80) nb+;else if(mark >=70) nc+;else if(mark >=60) nd+;else ne+; printf("Number of A(90-100): %dn", na); printf("Number of B(80-89): %dn", nb); printf("Number of C(70-79): %dn", nc); printf("Nu