数组典型例题及参考答案

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1、■第7章数组7-1输入一个正整数n（19510）,再输入n个整数（1）输出最大数和最小数，并输出平均值。（2）将最小数与第一个数交换，最大数与最后一个数交换，然后输出交换后的结果。（3）输出所有比平均值大的数。（4）找到最接近平均值的数。（提示：考虑差的绝对值）Program#defineN10main(){inta[N],max,min,sum,i,p,q,t;floatave,sub;/*输出最大数、最小数和平均数*/printf(”Pleaseinput%dintegers:'”；N);for(i=0;ivN;i++){scanf("%d",&a[i]);if(i

2、==0){max=min=sum=a[i];P=q=0;}else{if(maxa[i]){min=a[i];q=i；}sum=sum+a[i];}}ave=sum*1.0/N;printf(”Themaximumis:%d,theminimumis:%d,theaverageis%.2f”，max,min,ave);/*将最小数与第一个数交换，最大数与最后一个数交换*/t=a[O];a[O]=a[q];a[q]=t；t=a[N-1];a[N-1]=a[p];a[p]=t；printf(”Afterexchan

3、ge:”)；for(i=0;iave)printf(”％dH,a[i]);/*找到最接近平均值的数*/printf(”Thenumberclosesttotheaverageis:”)；sub=a[0]>ave?a[0]-ave:ave-a[0];P=0;for(i=1;iave?a[i]-ave:ave-a[i])

4、i]>ave?a[i]-ave:ave-a[i];P=i；}}printf(”％d.”,a[p]);}Output:Pleaseinput10integers::1734-1125107・9025J产input*/:Themaximumis:34,theminimumis:theaverageis8.00:Afterexchange:-11251725107・9034:Largerthantheaverage:25171034:Thenumberclosesttotheaverageis:77・2输入一个口期（按照year-month-day格式），计算该口期为当年的第

5、几天。（提示：注意闰年）（1）使用数组存储当年每个月应有的天数；（2）不使用数组完成；Programmain(){intm[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};intyear,month,day,i,total=0;printf(”Pleaseinputadate:(year-month-day)”)；scanf("%d-%d-%dn,&year,&month,&day);total=day;for(i=1;i

6、%400==0)&&month>=3)total++;printf("Thisdateisthe%dthdayintheyear.'n”,total);}Program(2)main(){intyear,month,day,total=0;printf(”Pleaseinputadate:(year-month-day)”);scanf(”％d・％d・％d”，&year,&month,&day);switch(month){case12:total+=30;case11:total+=31;case10:total+=30;case9:total+=31;case8:t

7、otal+=31;case7:total+=30;case6:total+=31;case5:total+=30;case4:total+=31;case3:total+=2&case2:total+=31;case1:total+=day;}if((year%4==0&&year%100!=0)11(year%400==0)&&month>=3)total++;printf(”Thisdateisthe%dthdayintheyear.H,total);Output:Pleaseinputadate:(year-month-day)