1、習(xí)題4參考解答1寫出下面函數(shù)的運(yùn)行結(jié)果。#include<stdio.h>voidfx701(intk,intn,floata);intmain()intk=1,n=5;floata=3.2,b=8.5;printf("調(diào)用前:k=%dn=%da=%.2fb=%.2fn",k,n,a,b);fx701(k,n,a);printf("調(diào)用后:k=%dn=%da=%.2fb=%.2fn",k,n,a,b);return0;voidfx701(intk,intn,floata)intq=100;floatb,x=6.2;k+=q;n*=q;a-=x

2、;b=3.5*a;printf("函數(shù)中:k=%dn=%da=%.2fb=%.2fn",k,n,a,b);解：調(diào)用前：k=1n=5a=3.20b=8.50函數(shù)中：k=101n=500a=-3.00b=-10.50調(diào)用后：k=1n=5a=3.20b=8.502編寫一個(gè)函數(shù)，實(shí)現(xiàn)從鍵盤輸入一個(gè)整數(shù)，如果該整數(shù)為素?cái)?shù)，則返回1，否則返0。解：#include<stdlib.h>#include<stdio.h>intIs_prime(inti)intj=0;for(j=2;j<i;j+)if(0=(i%j)return0;return1;intmai

3、n()inti;printf("Pleaseinputanumber:");scanf("%d",&i);if(1=Is_prime(i)printf("%disaprime.n",i);elseprintf("%disnotaprime.n",i);system("pause");return0;程序運(yùn)行結(jié)果如下：Pleaseinputanumber:13/13isaprime.Pleaseinputanumber:8/8isnotaprime.3設(shè)計(jì)兩個(gè)函數(shù)，分別求矩形的周長(zhǎng)和面積，

4、并在主函數(shù)中調(diào)用這兩個(gè)函數(shù)。解：#include<stdio.h>floatarea(floata,floatb)floats;s=a*b;return(s);floatzc(floata,floatb)floatp;p=2*(a+b);return(p);intmain()floata,b,s,p;printf("請(qǐng)輸入長(zhǎng)方形的邊長(zhǎng)a,b:");scanf("%f%f",&a,&b);s=area(a,b);p=zc(a,b);printf("長(zhǎng)方形的周長(zhǎng)為p=%f,長(zhǎng)方形的面積為s=%f",p,s);程

5、序運(yùn)行情況如下：請(qǐng)輸入長(zhǎng)方形的邊長(zhǎng)a,b:34/長(zhǎng)方形的周長(zhǎng)為p=14.000000,長(zhǎng)方形的面積為s=12.0000004編寫一個(gè)函數(shù)求一個(gè)整數(shù)10 ！的值。n的階乘，在主函數(shù)中計(jì)算并輸出解：#include<stdio.h>floatfc(intn)floats;if(n=1|n=0)s=1;elses=n*fc(n-1);return(s);intmain()floatk;k=fc(10);printf("10!=%fn",k);return0;程序運(yùn)行結(jié)果如下：10!=3628800.0000005已知函數(shù)fun2的原型說(shuō)明為“l(fā)ongfun2(intm

6、,intn);”，調(diào)用該函數(shù)計(jì)算mn的和(其中n>m>0),例如：若m=5,n=20,則計(jì)算5+6+7+8+-+20的值。要求在主函數(shù)中通過(guò)鍵盤輸入m和n的值，并輸出和的值。解：#include<stdio.h>longfun2(intm,intn)longs;inti;s=0;for(i=m;i<=n;i+)s=s+i;return(s);intmain()intn,m,sum;printf("請(qǐng)輸入m,n的值n>m:n");scanf("%d%d",&m,&n);sum=fun2(m,n);prin

7、tf("sum=%dn",sum);return0;程序運(yùn)行結(jié)果如下：請(qǐng)輸入m,n的值n>m:5 20/sum=2006 編寫一個(gè)函數(shù)重復(fù)打印給定的字符n次。解：#include<stdio.h>voidfun2(charm,intn)inti;for(i=1;i<=n;i+)printf("%c",m);printf("n");intmain()charm;intn;printf（"請(qǐng)輸入給定的字符n"）;scanf("%c",&m);printf("

8、請(qǐng)輸入打印的次數(shù)n");scanf("%d",&n);fun2(m,n);return0;程序運(yùn)行結(jié)果如下：請(qǐng)輸入給定的字符*/請(qǐng)輸入打印的次數(shù)5/*7寫出程序的運(yùn)行結(jié)果，并上機(jī)驗(yàn)證其結(jié)果。#include<stdio.h>voidfun1()intx=5;printf("x=%dn",x);voidfun2(intx)printf("x=%dn",+x);intmain()intx=2;fun1();fun2(x);printf("x=%dn",x);解：x=5x=3x=28指出下列

9、程序中各變量的存儲(chǔ)屬性，并寫出程序的執(zhí)行結(jié)果。( 1) #include<stdio.h>voidadd(inti);intmain()inti=1;add(i);printf("main:i=%dn",i);return0;voidadd(inti)i=i+3;printf("add:i=%dn",i);解：add:i=4main:i=1其中，i是自動(dòng)型變量。( 2) #include<stdio.h>inti=1;voidadd1();intmain()add1();printf("i=%dn",i);re

10、turn0;voidadd1()i=i+3;解：i=4其中，該變量是外部型變量。( 3) #include<stdio.h>voidadd2(inti);intmain()inti=2;add2(i);add2(i);return0;voidadd2(inti)staticintj=5;j=i+j;printf("j=%dn",j);解：j=7j=9其中，變量i是靜態(tài)型變量。9.編寫程序?qū)崿F(xiàn)以下圖形的輸出。*解:#include<stdio.h>voidmain()inti,j,k；for(i=1;i<=5;i+)for(j=1;j<=5

11、-i;j+)printf("");for(k=1;k<=2*i-1;k+)printf("*");printf("n");10.寫出程序的運(yùn)行結(jié)果。#include<stdio.h>先打印上邊的四行控制要打印的空格數(shù)量控制要打印的星號(hào)數(shù)# defineX5# defineYX+1# defineZY*X/2intmain()inta;a=Y;printf("%d%dn",Z,-a);return0;解：7511編寫一個(gè)函數(shù)，將一個(gè)十進(jìn)制整數(shù)轉(zhuǎn)換為二進(jìn)制數(shù)。解：#include<stdio.h

12、>voidmain()longnum;intozb();printf("請(qǐng)輸入一個(gè)十進(jìn)制數(shù)：");scanf("%ld",&num);printf("%ld的二進(jìn)制數(shù)為：",num);ozb(num);intozb(longnum)inta32=0,i=0;while(num>=1)ai=num%2;num=num/2;i+;for(i=i-1;i>=0;i-)printf("%d",ai);printf("n");return(ai);程序運(yùn)行結(jié)果如下：請(qǐng)輸入一個(gè)十進(jìn)制數(shù)：5/5的二進(jìn)制數(shù)為：10112編寫一個(gè)函數(shù)，找出任意整數(shù)的全部因子。解：#include<stdio.h&g