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ID:10271192
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页数:41页
时间:2018-06-14
《c语言程序设计作业整理ⅱ_浙大软院c语言》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、7.1.1输入一个正整数n(02、ncludeintmain(void){inti,j,n,x;intrepeat,ri;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;ia[i])continue;j=n-1;while(j>=i){a[j+1]=a[j];j-3、-;}a[i]=x;break;}if(i==n)a[n]=x;for(i=0;iintmain(void){inti,n,sum;intrepeat,r4、i;inta[10];doubleaver;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i5、),再输入n个整数,存入数组a中,先将数组a中的这n个数逆序存放,再按顺序输出数组中的n个元素。输入输出示例:括号内为说明输入2(repeat=2)4(n=4)108125(n=5)12540输出2181004521#includeintmain(void){inti,n,temp;intrepeat,ri;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i6、nf("%d",&a[i]);/*---------*/for(i=0;i<=(n-1)/2;i++){temp=a[i];a[i]=a[n-1-i];a[n-1-i]=temp;}for(i=0;i7、includeintmain(void){inti,index,n;intri,repeat;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
2、ncludeintmain(void){inti,j,n,x;intrepeat,ri;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;ia[i])continue;j=n-1;while(j>=i){a[j+1]=a[j];j-
3、-;}a[i]=x;break;}if(i==n)a[n]=x;for(i=0;iintmain(void){inti,n,sum;intrepeat,r
4、i;inta[10];doubleaver;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i5、),再输入n个整数,存入数组a中,先将数组a中的这n个数逆序存放,再按顺序输出数组中的n个元素。输入输出示例:括号内为说明输入2(repeat=2)4(n=4)108125(n=5)12540输出2181004521#includeintmain(void){inti,n,temp;intrepeat,ri;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i6、nf("%d",&a[i]);/*---------*/for(i=0;i<=(n-1)/2;i++){temp=a[i];a[i]=a[n-1-i];a[n-1-i]=temp;}for(i=0;i7、includeintmain(void){inti,index,n;intri,repeat;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
5、),再输入n个整数,存入数组a中,先将数组a中的这n个数逆序存放,再按顺序输出数组中的n个元素。输入输出示例:括号内为说明输入2(repeat=2)4(n=4)108125(n=5)12540输出2181004521#includeintmain(void){inti,n,temp;intrepeat,ri;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i6、nf("%d",&a[i]);/*---------*/for(i=0;i<=(n-1)/2;i++){temp=a[i];a[i]=a[n-1-i];a[n-1-i]=temp;}for(i=0;i7、includeintmain(void){inti,index,n;intri,repeat;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
6、nf("%d",&a[i]);/*---------*/for(i=0;i<=(n-1)/2;i++){temp=a[i];a[i]=a[n-1-i];a[n-1-i]=temp;}for(i=0;i7、includeintmain(void){inti,index,n;intri,repeat;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
7、includeintmain(void){inti,index,n;intri,repeat;inta[10];scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d",&n);for(i=0;i8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
8、4)for(i=1;ia[index])index=i;printf("max=%d,index=%d",a[index],index);}return0;}7.1.5输入一个正整数n(1
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