实验四答案资料

FILENAME实验四答案答案PAGEPAGE5实验四选择结构程序设计答案编程练习1：简单if语句的用法编写并调试程序：从键盘输入一个任意大小的实数x，如果满足如下条件：-29.5≤x≤29.5则输出x及其绝对值。（以simple_if.c文件名保存）源程序：#include<stdio.h>#include<math.h>voidmain(){doublex;printf("请输入一个任意大小的实数x\n");scanf("%lf",&x);if(x>=-29.5&&x<=29.5)printf("x=%lf,|x|=%lf\n",x,fabs(x));}编程练习2：if-else语句的用法修改练习1的程序，使得x满足-29.5≤x≤29.5时输出x及其绝对值，否则只输出x。（以if_else.c文件名保存）源程序：#include<stdio.h>#include<math.h>voidmain(){doublex;printf("请输入一个任意大小的实数x\n");scanf("%lf",&x);if((x>=-29.5)&&(x<=29.5))printf("x=%lf,|x|=%lf\n",x,fabs(x));elseprintf("x=%lf\n",x);}编程练习3：if-elseif语句的用法编写一个能够进行加减法运算的程序，要求：从键盘任意输入两个实数，然后输入一个运算符，当运算符为“+”时对两个数进行加法运算，为“-”时对两个数进行减法运算，其它符号时，只显示输入的符号不进行运算。（以if_else_if.c文件名保存）源程序：(方法1：else子句中嵌套if-else结构)#include<stdio.h>voidmain(){doublex,y;charch;printf("请输入任意两个实数和一个数学运算符(如:34.5+63.9)\n");scanf("%lf%c%lf",&x,&ch,&y);if(ch!='+'&&ch!='-')printf("输入的运算符为%c号,不进行运算\n",ch);elseif(ch=='+')printf("输入的运算符为%c号,运算结果是:%lf+%lf=%lf\n",ch,x,y,x+y); elseprintf("输入的运算符为%c号,运算结果是:%lf-%lf=%lf\n",ch,x,y,x-y);}源程序：(方法2：if子句中嵌套if-else结构)#include<stdio.h>voidmain(){doublex,y;charch;printf("请输入任意两个实数和一个数学运算符(如:34.5+63.9)\n");scanf("%lf%c%lf",&x,&ch,&y);if(ch=='+'||ch=='-')if(ch=='+')printf("输入的运算符为%c号,运算结果是:%lf+%lf=%lf\n",ch,x,y,x+y); elseprintf("输入的运算符为%c号,运算结果是:%lf-%lf=%lf\n",ch,x,y,x-y);elseprintf("输入的运算符为%c号,不进行运算\n",ch);}源程序：(方法3：使用switch语句，虽不符合练习要求，但对于此问题的确是一种好方法！)#include<stdio.h>voidmain(){doublex,y;charch;printf("请输入任意两个实数和一个数学运算符(如:34.5+63.9)\n");scanf("%lf%c%lf",&x,&ch,&y);if(m==11)days=31+28+31+30+31+30+31+31+30+31+d;if(m==12)days=31+28+31+30+31+30+31+31+30+31+30+d;}printf("\n%d-%d-%disNo.%d.\n",y,m,d,days);}源程序：(方法2)#include<stdio.h>main(){intday,month,year,sum,leap;printf("Pleaseinputdate(yyyy-mm-dd):\n");a1:scanf("%d-%d-%d",&year,&month,&day);if(year<=0||year>10000||month<=0||month>12||day<=0||day>31){printf("dataerror!pleaseinputagain:\n");gotoa1;}switch(month)/*先计算某月以前月份的总天数*/{case1:sum=0;break;case2:sum=31;break;case3:sum=59;break;case4:sum=90;break;case5:sum=120;break;case6:sum=151;break;case7:sum=181;break;case8:sum=212;break;case9:sum=243;break;case10:sum=273;break;case11:sum=304;break;case12:sum=334;break;default:printf("dataerror");break;}sum=sum+day;/*再加上某天的天数*/if(year%400==0||(year%4==0&&year%100!=0))/*判断是不是闰年*/leap=1;elseleap=0;if(leap==1&&month>2)/*如果是闰年且月份大于2,总天数应该加一天*/sum++;printf("Itisthe%dthday.\n",sum);}源程序：(方法3：需要用到循环语句)#include"stdio.h"main(){inti,y,m,d,days;printf("Pleaseinputdate(yyyy-mm-dd):\n");a1:scanf("%d-%d-%d",&y,&m,&d);if(y<0||y>10000||m<0||m>12||d<0||d>31){printf("dataerror!pleaseinputagain:\n");gotoa1;}days=d;if(y%4==0&&y%100!=0||y%400==0)for(i=1;i<m;i++)if(i==4||i==6||i==9||i==11)days=days+30; elseif(i==2)days=days+29; elsedays=days+31;elsefor(i=1;i<m;i++)if(i==4||i==6||i==9||i==11)days=days+30; elseif(i==2)days=days+28; elsedays=days+31;printf("\n%d-%d-%disNo.%d.\n",y,m,d,days);}(4)求解任意的一元二次方程ax2+bx+c=0的根，a,b,c的值由键盘输入（考虑解的各种情况）。源程序：#include<stdio.h>#include<math.h>voidmain(){

floata,b,c,p,x1,x2;

printf("请输入一元二次方程的三个系数a,b,c的值:\n");

scanf("%f%f%f",&a,&b,&c);

p=b*b-4*a*c;if(p<0)

printf("此方程没有实根!\n");

elseif(p==0)

printf("此方程有两个相等的实根:%f和%f\n",-b/(2*a),-b/(2*a));

else

printf("此方程有两个不等的实根:%f和%f!\n",(-b+sqrt(p))/(2*a),(-b-sqrt(p))/(2*a));}(5)一个3位数，如果它的各位数字之的立方和等于该数本身，则称其为水仙花数。源程序：#include<stdio.h>#include<math.h>voidmain(){intnum,a,b,c;printf(“请任意输入一个三位正整数\n”);scanf(“%d”,&num);a=num/100;b=num/10%10;c=num%10;if(num==pow(a,3)+pow(b,3)+pow(c,3))

printf("%d是水仙花数。\n",num);elseprintf("%d不是水仙花数。\n",num);}注意：三位的水仙花数有153、370、371、407学了循环以后如何编程呢附加编程题(1)给出一个不多于3位的正整数，编写程序：①求出它是几位数。②分别打印出每一位数字。③按反序打印出每位数字。例如：原数是123，应输出321。源程序：(方法1：使用三个并列的if结构)#include"stdio.h"main(){intn;printf("请输入一个不多三位的正整数\n");label:scanf("%d",&n);if(n<=0||n>999){printf("数据错误！请重新输入：");gotolabel;}if(n>=1&&n<=9){ printf("\n此数是一位数，");printf("个位数字是%d",n);printf("\n反序数字为%d\n",n);}if(10<=n&&n<=99){ printf("\n此数是两位数，");printf("十位数字是%d,",n/10); printf("个位数字是%d。",n%10);printf("\n反序数字为%d%d\n",n%10,n/10);}if(100<=n&&n<=999){ printf("\n此数是三位数，"); printf("百位数字是%d，",n/100); printf("十位数字是%d，",n/10%10); printf("个位数字是%d。",n%10);printf("\n反序数字为%d%d%d\n",n%10,n/10%10,n/100);}}源程序：(方法2：使用if-elseif结构)#include"stdio.h"main(){intn;printf("请输入一个不多三位的正整数\n");label:scanf("%d",&n);if(n<=0||n>999){printf("数据错误！请重新输入：");gotolabel;}if(n>=1&&n<=9){ printf("\n此数是一位数，");printf("个位数字是%d",n);printf("\n反序数字为%d\n",n);}elseif(n<=99){ printf("\n此数是两位数，");printf("十位数字是%d,",n/10); printf("个位数字是%d。",n%10);printf("\n反序数字为%d%d\n",n%10,n/10);}elseif(n<=999){ printf("\n此数是三位数，"); printf("百位数字是%d，",n/100); printf("十位数字是%d，",n/10%10); printf("个位数字是%d。",n%10);printf("\n反序数字为%d%d%d\n",n%10,n/10%10,n/100);}}(2)编程解决与日历有关的问题：①输入年、月，求该月的天数。②给出年、月、日，计算出该日是该年的第几天。如输入2000，3，1，则输出61。③1990年元旦是星期一，问1900年7月1日是星期几。①源程序：#include"stdio.h"main(){inty,m,d;printf("Pleaseinputyear-month:\n");a1:scanf("%d-%d",&y,&m);if(y<=0||y>10000||m<=0||m>12){printf("dataerror!pleaseinputagain:\n");gotoa1;}if(m==2)if(y%4==0&&y%100!=0||y%400==0) d=29;elsed=28;elseif(m==4||m==6||m==9||m==11)d=30;elsed=31;printf("%d-%dis%ddays.\n",y,m,d);}②源程序：(方法1)#include"stdio.h"main(){inty,m,d,days;printf("Pleaseinputdate(yyyy-mm-dd):\n");a1:scanf("%d-%d-%d",&y,&m,&d);if(y<=0||y>10000||m<=0||m>12||d<=0||d>31){printf("dataerror!pleaseinputagain:\n");gotoa1;}if(y%4==0&&y%100!=0||y%400==0){if(m==1)days=d;if(m==2)days=31+d;if(m==3)days=31+29+d;if(m==4)days=31+29+31+d;if(m==5)days=31+29+31+30+d;if(m==6)days=31+29+31+30+31+d;if(m==7)days=31+29+31+30+31+30+d;if(m==8)days=31+29+31+30+31+30+31+d;if(m==9)days=31+29+31+30+31+30+31+31+d;if(m==10)days=31+29+31+30+31+30+31+31+30+d;if(m==11)days=31+29+31+30+31+30+31+31+30+31+d;if(m==12)days=31+29+31+30+31+30+31+31+30+31+30+d;}else{if(m==1)days=d;if(m==2)days=31+d;if(m==3)days=31+28+d;if(m==4)days=31+28+31+d;if(m==5)days=31+28+31+30+d;if(m==6)days=31+28+31+30+31+d;if(m==7)days=31+28+31+30+31+30+d;if(m==8)days=31+28+31+30+31+30+31+d;if(m==9)days=31+28+31+30+31+30+31+31+d;if(m==10)days=31+28+31+30+31+30+31+31+30+d;if(m==11)days=31+28+31+30+31+30+31+31+30+31+d;if(m==12)days=31+28+31+30+31+30+31+31+30+31+30+d;}printf("\n%d-%d-%disNo.%d.\n",y,m,d,days);}源程序：(方法2)#include<stdio.h>main(){intday,month,year,sum,leap;printf("Pleaseinputdate(yyyy-mm-dd):\n");a1:scanf("%d-%d-%d",&year,&month,&day);if(year<=0||year>10000||month<=0||month>12||day<=0||day>31){printf("dataerror!pleaseinputagain:\n");gotoa1;}switch(month)/*先计算某月以前月份的总天数*/{case1:sum=0;break;case2:sum=31;break;case3:sum=59;break;case4:sum=90;break;case5:sum=120;break;case6:sum=151;break;case7:sum=181;break;case8:sum=212;break;case9:sum=243;break;case10:sum=273;break;case11:sum=304;break;case12:sum=334;break;default:printf("dataerror");break;}sum=sum+day;/*再加上某天的天数*/if(year%400==0||(year%4==0&&year%100!=0))/*判断是不是闰年*/leap=1;elseleap=0;if(leap==1&&month>2)/*如果是闰年且月份大于2,总天数应该加一天*/sum++;printf("Itisthe%dthday.\n",sum);}源程序：(方法3：需要用到循环语句)#include"stdio.h"main(){inti,y,m,d,days;printf("Pleaseinputdate(yyyy-mm-dd):\n");a1:scanf("%d-%d-%d",&y,&m,&d);if(y<0||y>10000||m<0||m>12||d<0||d>31){printf("dataerror!pleaseinputagain:\n");gotoa1;}days=d;if(y%4==0&&y%100!=0||y%400==0)for(i=1;i<m;i++)if(i==4||i==6||i==9||i==11)days=days+30; elseif(i==2)days=days+29; elsedays=days+31;elsefor(i=1;i<m;i++)if(i==4||i==6||i==9||i==11)days=days+30; elseif(i==2)days=days+28; elsedays=days+31;printf("\n%d-%d-%disNo.%d.\n",y,m,d,days);}③源程序：#include"stdio.h"main(){inty=1900,d1=1,d2=1,days,xingqi;if(y%4==0&&y%100!=0||y%400==0)/*if结构求1900年7days=31+29+31+30+31+30+d2;elsedays=31+28+31+30+31+30+d2;xingqi=1+(days-d1)%7;/*先求1900年7月1日与1900年printf("\n1900年7if(xingqi==1)printf("一。\n");if(xingqi==2)printf("二。\n");if(xingqi==3)printf("三。\n");if(xingqi==4)printf("四。\n");if(xingqi==5)printf("五。\n");if(xingqi==6)printf("六。\n");if(xingqi==7)printf("日。\n");}思维扩展：求任意年任意月任意日是星期几。源程序：run(intyear){intr=0;if((year%4==0&&year%100!=0)||(year%100==0&&year%400==0))r=1;return(r);}longdays(intyear,intmonth,intday){inta[12]={31,28,31,30,31,30,31,31,30,31,30,31};inti,b,x,y;longz;i=b=x=y=0;for(i=1;i<year;i++){b=run(i);if(b)x++;}b=run(year);for(i=1;i<month;i++)y+=a[i-1];if(month>2&&b)y=y+1;z=365l*(year-1)+y+(day-1)+x;return(z);}longxingqi(intyear,intmonth,intday){longa,b,c;a=days(1900,1,1);b=days(year,month,day);c=1+((b-a)%7+7)%7;return(c);}main(){longa,b,c