c语言大作业仅供参考

1、实验综合练打印万年历(1)已知公元 1 年 1 月 1 日是编程如下：#include void main()一，计算输入的年月日是几？i,year,th,day,day_of_year,day_of_th;long sumday=0;prf(请先输入年、月、日，中间以空格或换行符间隔：n);scanf(%ld%ld%ld,&year,&for(i=1;i=year-1;i+)th,&day);if(i%400=0|i%100!=0&i%4=0)day_of_year=366;elseday_of_year=365; sumday=sumday+day_of_year;for(i=1;i=th

2、-1;i+)if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)day_of_th=31;else if(i=4|i=6|i=9|i=11)day_of_else if(i=2)th=30;if(year%400=0|year%100!=0&year%4=0)day_of_th=29;elseday_of_th=28;sumday=sumday+day_of_th;sumday=sumday+day; i=sumday%7;if(i=0)i=7;prf(%ld 年%ld 月%ld 日是%dn,year,th,day,i);为了使输出的为大写且将#include void main

3、()7 变成天，可作如下改动：i,year,th,day,day_of_year,day_of_th;long sumday=0;prf(请先输入年、月、日，中间以空格或换行符间隔：n);scanf(%ld%ld%ld,&year,&for(i=1;i=year-1;i+)th,&day);if(i%400=0|i%100!=0&i%4=0)day_of_year=366;elseday_of_year=365; sumday=sumday+day_of_year;for(i=1;i=th-1;i+)if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)day_of_th=31;e

4、lse if(i=4|i=6|i=9|i=11)day_of_else if(i=2)th=30;if(year%400=0|year%100!=0&year%4=0)day_of_th=29;elseday_of_th=28;sumday=sumday+day_of_th;sumday=sumday+day; i=sumday%7; switch(i)pr f(%ld 年%ld 月%ld 日是 pr f(%ld 年%ld 月%ld 日是 pr f(%ld 年%ld 月%ld 日是 pr f(%ld 年%ld 月%ld 日是 pr f(%ld 年%ld 月%ld 日是 pr f(%ld 年%l

5、d 月%ld 日是 pr f(%ld 年%ld 月%ld 日是日,year,一,year,二,year,三,year,四,year,五,year,六,year,case 0:case 1:case 2:case 3:case 4:case 5:case 6:th,day); break;th,day); break; th,day); break; th,day); break; th,day); break; th,day); break; th,day); break;prf(n);当然也可以这样编程：#include leap_year(year)if(year%400=0 | year

6、%100!=0 & year%4=0)return 1;elsereturn 0;void main()long year,th, day;char *a7=日,六;一,二,三,四,五,prf(请先输入年、月份、日，中间以空格或换行符间隔：n);scanf(%ld%ld%ld,&year,&i, xq=0,sumday=0; for(i=1;i=year-1;i+)th,&day);sumday=sumday+365+leap_year(i);for(i=1;i=th-1;i+)if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)sumday=sumday+31;else if(

7、i=4|i=6|i=9|i=11) sumday=sumday+30;else if(i=2) sumday=sumday+28+leap_year(year);sumday=sumday+day; i=sumday%7;prf(%ld 年%ld 月%ld 日是%sn,year,th,day,ai);(2)输入年份，在屏幕上输出该年的年历（已知 1900 年第一天也是#include void main()一）。year,i,j,k,sumday,yuandan,f(Please input the year:n);kday,day,day_of_year,day_of_th;prscanf(

8、%d,&year);prf(n);for(i=1,sumday=0;i=year-1990;i+)if(i%400=0|i%100!=0&i%4=0) day_of_year=366;elseday_of_year=365;sumday=sumday+day_of_year;yuandan=(sumday+1)%7; kday=yuandan;f(公元%d 年n,year);prfor(i=1;i=12;i+)f(%d 月n,i);prprf(SunTue Wed Thu Fri Satn);for(j=0;jkday;j+)prf();if(i=1|i=3|i=5|i=7|i=8|i=10|

9、i=12)day_of_th=31;else if(i=4|i=6|i=9|i=11)day_of_else if(i=2)th=30;if(year%400=0 | year%100!=0 & year%4=0)day_of_th=29;elseday_of_th=28;day=day_of_th;for(k=1;k=day;k+)prf(%4d,k);kday=(kday+1)%7;if(kday=0)prf(n);prf(nn);#include void main()year,i,j,k,sumday,yuandan,kday,day,day_of_year,day_of_th;prf

10、(Please input the year:n);scanf(%d,&year);prf(n);for(i=1,sumday=0;i=year-1;i+)if(i%400=0|i%100!=0&i%4=0)day_of_year=366;elseday_of_year=365; sumday=sumday+day_of_year;yuandan=(sumday+1)%7; kday=yuandan;f(公元%d 年n,year);prfor(i=1;i=12;i+)switch(i)case 1:prcase 2:pr case 3:pr case 4:pr case 5:pr case 6

11、:pr case 7:pr case 8:pr case 9:pr case 10:pr case 11:pr case 12:prf(f(f(f(f(f(f(f(f(f(f(f(Januaryn); break;Februaryn); break; Marchn); break; Apriln); break; Mayn); break; Junen); break; Julyn); break; Augustn); break; Septembern); break; Octobern); break; Novembern); break; Decembern);break;prf(Sun

12、Tue Wed Thu Fri Satn);for(j=0;jkday;j+)prf();if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)day_of_th=31;else if(i=4|i=6|i=9|i=11)day_of_else if(i=2)th=30;if(year%400=0 | year%100!=0 & year%4=0)day_of_th=29;elseday_of_th=28;day=day_of_th;for(k=1;k=day;k+)prf(%4d,k);kday=(kday+1)%7;if(kday=0)prf(n);prf(n);或者: #inc

13、lude void main()year,i,j,k,sumday,yuandan,kday,day,day_of_year,day_of_th;cha*a13=,January,February,March,April,May,June,July,August,September,October,November,December;prf(Please input the year:n);scanf(%d,&year);prf(n);for(i=1,sumday=0;i=year-1990;i+)if(i%400=0|i%100!=0&i%4=0) day_of_year=366;elsed

14、ay_of_year=365; sumday=sumday+day_of_year;yuandan=(sumday+1)%7; kday=yuandan;f(公元%d 年n,year);prfor(i=1;i=12;i+)prprf(%sn”,ai);f(SunTue Wed Thu Fri Satn);for(j=0;jkday;j+)prf();if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)day_of_th=31;else if(i=4|i=6|i=9|i=11)day_of_else if(i=2)th=30;if(year%400=0 | year%100!=0

15、& year%4=0)day_of_th=29;elseday_of_th=28;day=day_of_th;for(k=1;k=day;k+)prf(%4d,k);kday=(kday+1)%7;if(kday=0)prf(n);prf(nn);当然也可以定义函数和调用定义的函数：#includeleap_year(year)if(year%400=0 | year%100!=0 & year%4=0)return 1;elsereturn 0;day_of_th(i,year)if(i=1|i=3|i=5|i=7|i=8|i=10|i=12)return 31;else if(i=4|i=

16、6|i=9|i=11) return 30;else if(i=2)return 28+leap_year(year);void main()year,i,j,k,sumday,day,kday,day;char*a13=,January,February,March,April,May,June,July,August,September,October,November,December;prf(Please input the year:n);scanf(%d,&year);prf(n);for(i=1,sumday=0;i=year-1990;i+)sumday=sumday+365+

17、leap_year(i); day=(sumday+1)%7;kday=day;f(%d 年n,year);prfor(i=1;i=12;i+)prprf(%sn,ai);f(SunTue Wed Thu Fri Satn);for(j=0;jkday;j+)prf();th(i,year);day=day_of_for(k=1;k=day;k+)prf(%4d,k);kday=(kday+1)%7;if(kday=0)prf(n);prf(n);运行结果实例如下：2001JanuarySunTue29Wed310Thu411Fri512Sat613187February实验综合练习二（1）输

18、入一个班10 个学生的学号和每个学生三门功课（数、计算机基础）的成绩。编程计算出每个学生的总分和平均分，并按学生成绩优劣排序，最后打印一张按高分到低分名次排序的成绩表。要求：打印的成绩单表项包括：序号，学号、数编程如下：#include void main()i,j,k,sum,temp;a117;for(i=1,j=0;i11;i+) ai0=i;for(i=1;i11;i+)sum=0;prf(请输入第%d 名学生的学号:n,i);、计算机、总分、平均分。scanf(%d,&ai1);prf(请输入第%d 名学生的数学成绩:n,i); scanf(%d,&ai2);prf(请输入第%d 名

19、学生的英语成绩:n,i); scanf(%d,&ai3);prf(请输入第%d 名学生的计算机成绩:n,i); scanf(%d,&ai4);sum=sum+ai2+ai3+ai4; ai5=sum;ai6=ai5/3;for(i=1;i11;i+)for(k=1;k11-i;k+)if(ak5ak+15)for(j=1;j7;j+)temp=akj; akj=ak+1j; ak+1j=temp;f(序号 学号数学 英语计算机 总分 平均分n);prfor(i=1;i11;i+)for(j=0;j7;j+)prf(%6d,aij);prf(n);这里排序应用了冒泡排序，当然也可以应用选择排序，

20、自己可以去试一下！这里给大家一组数据：序号学号数计算机总分平均分44787593558487836682839077798791运行后的结果应该为：（2）输入一个班 10 个学生的学号和每个学生三门功课（数、计算机基础）的成绩。编程计算出每个学生的总分和平均分，并按学生成绩优劣排序，最后打印一张按高分到低分名次排序的成绩表。要求：1)排序用一个函数实现。2)打印的成绩单表项包括：序号，学号、数3）用结构体来存放学生的信息编程如下：#include struct student、计算机、总分、平均分。xuhao,stu_nuputer,sum,aver;void paixu(student *stu,n)student stud; i,j;for(i=0;in-1; i+) for(j=i+1;jn;j+)if(stui.sum stuj.sum)stud = stui; stui = stuj; stuj = stud;void main( )序号学号数计算机总分平均分8921878592264883985869