杭电题目acm答案

./选修课考试作业TOC\o"1-1"\h\z\u1001SumProblem21089A+BforInput-OutputPractice<I>41090A+BforInput-OutputPractice<II>61091A+BforInput-OutputPractice<III>81092A+BforInput-OutputPractice<IV>91093A+BforInput-OutputPractice<V>111094A+BforInput-OutputPractice<VI>121095A+BforInput-OutputPractice<VII>131096A+BforInput-OutputPractice<VIII>142000ASCII码排序162001计算两点间的距离172002计算球体积192003求绝对值202004成绩转换212005第几天？222006求奇数的乘积242007平方和与立方和262008数值统计272009求数列的和282010水仙花数292011多项式求和312012素数判定332014青年歌手大奖赛_评委会打分342015偶数求和362016数据的交换输出382017字符串统计402019数列有序!412020绝对值排序432021发工资咯：452033人见人爱A+B462039三角形482040亲和数49：春杰班级：电商1001学号：1010504134.1001SumProblemProblemDescriptionHey,welcometoHDOJ<HangzhouDianziUniversityOnlineJudge>.

Inthisproblem,yourtaskistocalculateSUM<n>=1+2+3+...+n.InputTheinputwillconsistofaseriesofintegersn,oneintegerperline.OutputForeachcase,outputSUM<n>inoneline,followedbyablankline.Youmayassumetheresultwillbeintherangeof32-bitsignedinteger.SampleInput1100SampleOutput15050AuthorDOOMIII解答：#include<stdio.h>main<>{intn,i,sum;sum=0;while<<scanf<"%d",&n>!=-1>>{sum=0;for<i=0;i<=n;i++>sum+=i;printf<"%d\n\n",sum>;}}.1089A+BforInput-OutputPractice<I>ProblemDescriptionYourtaskistoCalculatea+b.

Tooeasy?!Ofcourse!Ispeciallydesignedtheproblemforacmbeginners.

Youmusthavefoundthatsomeproblemshavethesametitleswiththisone,yes,alltheseproblemsweredesignedforthesameaim.InputTheinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.OutputForeachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.SampleInput151020SampleOutput630AuthorlcyRecommendJGShining解答：#include<stdio.h>main<>{inta,b;while<scanf<"%d%d",&a,&b>!=EOF>printf<"%d\n",a+b>;}.1090A+BforInput-OutputPractice<II>ProblemDescriptionYourtaskistoCalculatea+b.InputInputcontainsanintegerNinthefirstline,andthenNlinesfollow.Eachlineconsistsofapairofintegersaandb,separatedbyaspace,onepairofintegersperline.OutputForeachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.SampleInput2151020SampleOutput630AuthorlcyRecommendJGShining解答：#include<stdio.h>#defineM1000voidmain<>{inta,b,n,j[M],i;//printf<"pleaseinputn:\n">;scanf<"%d",&n>;for<i=0;i<n;i++>{scanf<"%d%d",&a,&b>;//printf<"%d%d",a,b>;j[i]=a+b;}i=0;while<i<n>{printf<"%d",j[i]>;i++;printf<"\n">;}}.1091A+BforInput-OutputPractice<III>ProblemDescriptionYourtaskistoCalculatea+b.InputInputcontainsmultipletestcases.Eachtestcasecontainsapairofintegersaandb,onepairofintegersperline.Atestcasecontaining00terminatestheinputandthistestcaseisnottobeprocessed.OutputForeachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.SampleInput15102000SampleOutput630AuthorlcyRecommendJGShining解答：#include<stdio.h>main<>{inta,b;scanf<"%d%d",&a,&b>;while<!<a==0&&b==0>>{printf<"%d\n",a+b>;scanf<"%d%d",&a,&b>;}}1092A+BforInput-OutputPractice<IV>ProblemDescriptionYourtaskistoCalculatethesumofsomeintegers.InputInputcontainsmultipletestcases.EachtestcasecontainsaintegerN,andthenNintegersfollowinthesameline.Atestcasestartingwith0terminatestheinputandthistestcaseisnottobeprocessed.OutputForeachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput.SampleInput412345123450SampleOutput1015AuthorlcyRecommendJGShining解答：#include<stdio.h>intmain<>{intn,sum,i,t;while<scanf<"%d",&n>!=EOF&&n!=0>{sum=0;for<i=0;i<n;i++>{scanf<"%d",&t>;sum=sum+t;}printf<"%d\n",sum>;}}.1093A+BforInput-OutputPractice<V>ProblemDescriptionYourtaskistocalculatethesumofsomeintegers.InputInputcontainsanintegerNinthefirstline,andthenNlinesfollow.EachlinestartswithaintegerM,andthenMintegersfollowinthesameline.OutputForeachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput.SampleInput241234512345SampleOutput1015Authorlcy解答：#include<stdio.h>main<>{intn,a,b,i,j,sum;sum=0;while<scanf<"%d\n",&n>!=-1>{for<i=0;i<n;i++>{scanf<"%d",&b>;for<j=0;j<b;j++>{scanf<"%d",&a>;sum+=a;}printf<"%d\n",sum>;sum=0;}}}.1094A+BforInput-OutputPractice<VI>ProblemDescriptionYourtaskistocalculatethesumofsomeintegers.InputInputcontainsmultipletestcases,andonecaseoneline.EachcasestartswithanintegerN,andthenNintegersfollowinthesameline.OutputForeachtestcaseyoushouldoutputthesumofNintegersinoneline,andwithonelineofoutputforeachlineininput.SampleInput41234512345SampleOutput1015AuthorlcyRecommendJGShining解答：#include<stdio.h>main<>{intn,a,b,i,j,sum;sum=0;while<scanf<"%d\n",&n>!=-1>{for<j=0;j<n;j++>{scanf<"%d",&a>;sum+=a;}printf<"%d\n",sum>;sum=0;}}[CopytoClipboard][SavetoFile]1095A+BforInput-OutputPractice<VII>ProblemDescriptionYourtaskistoCalculatea+b.InputTheinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.OutputForeachpairofinputintegersaandbyoushouldoutputthesumofaandb,andfollowedbyablankline.SampleInput151020SampleOutput630AuthorlcyRecommendJGShining解答：#include<stdio.h>main<>{inta,b;while<scanf<"%d%d",&a,&b>!=EOF>printf<"%d\n\n",a+b>;}1096A+BforInput-OutputPractice<VIII>ProblemDescriptionYourtaskistocalculatethesumofsomeintegers.InputInputcontainsanintegerNinthefirstline,andthenNlinesfollow.EachlinestartswithaintegerM,andthenMintegersfollowinthesameline.OutputForeachgroupofinputintegersyoushouldoutputtheirsuminoneline,andyoumustnotethatthereisablanklinebetweenoutputs.SampleInput3412345123453123SampleOutput10156AuthorlcyRecommendJGShining解答：intmain<>{inta,b,i,j,l[1000],k;scanf<"%d",&i>;getchar<>;for<j=1;j<=i;j++>l[j]=0;for<j=1;j<=i;j++>{scanf<"%d",&a>;getchar<>;for<k=1;k<=a;k++>{scanf<"%d",&b>;getchar<>;l[j]+=b;}}for<j=1;j<=i-1;j++>printf<"%d\n\n",l[j]>;printf<"%d\n",l[i]>;}2000ASCII码排序ProblemDescription输入三个字符后,按各字符的ASCII码从小到大的顺序输出这三个字符。Input输入数据有多组,每组占一行,有三个字符组成,之间无空格。Output对于每组输入数据,输出一行,字符中间用一个空格分开。SampleInputqweasdzxcSampleOutputeqwadscxzAuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>main<>{chara,b,c,d;while<scanf<"%c%c%c",&a,&b,&c>!=EOF>{getchar<>;if<a>=b>{if<c>=a>printf<"%c%c%c\n",b,a,c>;elseif<b>=c>printf<"%c%c%c\n",c,b,a>;elseif<b<c>printf<"%c%c%c\n",b,c,a>;}else{if<c>=b>printf<"%c%c%c\n",a,b,c>;elseif<c>=a>printf<"%c%c%c\n",a,c,b>;elseif<a>c>printf<"%c%c%c\n",c,a,b>;}}}2001计算两点间的距离ProblemDescription输入两点坐标〔X1,Y1,〔X2,Y2,计算并输出两点间的距离。Input输入数据有多组,每组占一行,由4个实数组成,分别表示x1,y1,x2,y2,数据之间用空格隔开。Output对于每组输入数据,输出一行,结果保留两位小数。SampleInput00010110SampleOutput1.001.41AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>#include<math.h>main<>{doublea,b,c,d,s;while<scanf<"%lf%lf%lf%lf",&a,&b,&c,&d>!=EOF>{s=sqrt<<a-c>*<a-c>+<b-d>*<b-d>>;printf<"%.2lf\n",s>;}}2002计算球体积ProblemDescription根据输入的半径值,计算球的体积。Input输入数据有多组,每组占一行,每行包括一个实数,表示球的半径。Output输出对应的球的体积,对于每组输入数据,输出一行,计算结果保留三位小数。SampleInput11.5SampleOutput4.18914.137Hint#definePI3.1415927AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>#definePI3.1415927main<>{doublea,v;while<scanf<"%lf",&a>!=EOF>{v=4*PI*a*a*a/3;printf<"%.3lf\n",v>;}}2003求绝对值ProblemDescription数的绝对值。Input输入数据有多组,每组占一行,每行包含一个实数。Output对于每组输入数据,输出它的绝对值,要求每组数据输出一行,结果保留两位小数。SampleInput123-234.00SampleOutput123.00234.00AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>main<>{doublea;while<scanf<"%lf",&a>!=EOF>{if<a<0>a=-a;printf<"%.2lf\n",a>;}}2004成绩转换ProblemDescription输入一个百分制的成绩t,将其转换成对应的等级,具体转换规则如下：

90~100为A;

80~89为B;

70~79为C;

60~69为D;

0~59为E;Input输入数据有多组,每组占一行,由一个整数组成。Output对于每组输入数据,输出一行。如果输入数据不在0~100围,请输出一行："Scoreiserror!"。SampleInput5667100123SampleOutputEDAScoreiserror!AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>intmain<>{intn;while<scanf<"%d",&n>!=EOF>{if<n>100||n<0>printf<"Scoreiserror!\n">;elseif<n>=90>printf<"A\n">;elseif<n>=80>printf<"B\n">;elseif<n>=70>printf<"C\n">;elseif<n>=60>printf<"D\n">;elseprintf<"E\n">;}return0;}2005第几天？ProblemDescription给定一个日期,输出这个日期是该年的第几天。Input输入数据有多组,每组占一行,数据格式为YYYY/MM/DD组成,具体参见sampleinput,另外,可以向你确保所有的输入数据是合法的。Output对于每组输入数据,输出一行,表示该日期是该年的第几天。SampleInput1985/1/202006/3/12SampleOutput2071AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>main<>{inta,b,c,d,e,f,g;while<scanf<"%d/%d/%d",&a,&b,&c>!=EOF>{if<b==1>d=c;elseif<b==2>d=31+c;elseif<b==3>d=31+28+c;elseif<b==4>d=31+28+31+c;elseif<b==5>d=31+31+28+30+c;elseif<b==6>d=31+28+31+30+31+c;elseif<b==7>d=31+28+31+30+31+30+c;elseif<b==8>d=31+28+31+30+31+30+31+c;elseif<b==9>d=31+28+31+30+31+30+31+31+c;elseif<b==10>d=31+28+31+30+31+30+31+31+30+c;elseif<b==11>d=31+28+31+30+31+30+31+31+30+31+c;elseif<b==12>d=31+28+31+30+31+30+31+31+30+31+c+30;e=a%100;f=a%400;g=a%4;if<e==0>{if<f==0>d=1+d;elsed=d;}elseif<g=0>d=d+1;elsed=d;printf<"%d\n",d>;}}2006求奇数的乘积ProblemDescription给你n个整数,求他们中所有奇数的乘积。Input输入数据包含多个测试实例,每个测试实例占一行,每行的第一个数为n,表示本组数据一共有n个,接着是n个整数,你可以假设每组数据必定至少存在一个奇数。Output输出每组数中的所有奇数的乘积,对于测试实例,输出一行。SampleInput312342345SampleOutput315AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>main<>{intn,s,i,a;while<scanf<"%d",&n>!=EOF>{s=1;for<i=0;i<n;i++>{scanf<"%d",&a>;if<a%2==1>s=s*a;else;}printf<"%d\n",s>;}}2007平方和与立方和ProblemDescription给定一段连续的整数,求出他们中所有偶数的平方和以及所有奇数的立方和。Input输入数据包含多组测试实例,每组测试实例包含一行,由两个整数m和n组成。Output对于每组输入数据,输出一行,应包括两个整数x和y,分别表示该段连续的整数中所有偶数的平方和以及所有奇数的立方和。

你可以认为32位整数足以保存结果。SampleInput1325SampleOutput42820152AuthorlcySourceC语言程序设计练习〔一RecommendJGShining解答：#include<stdio.h>intmain<>{intsum1,sum2,n,i,m,t;while<scanf<"%d%d",&m,&n>!=EOF>{sum1=sum2=0;if<m>n>{t=m;m=n;n=t;}for<i=m;i<=n;i++>{if<i%2==0>sum1+=<i*i>;elsesum2+=<i*i*i>;}printf<"%d%d\n",sum1,sum2>;}return0;}2008数值统计ProblemDescription统计给定的n个数中,负数、零和正数的个数。Input输入数据有多组,每组占一行,每行的第一个数是整数n〔n<100,表示需要统计的数值的个数,然后是n个实数；如果n=0,则表示输入结束,该行不做处理。Output对于每组输入数据,输出一行a,b和c,分别表示给定的数据中负数、零和正数的个数。SampleInput60123-10512340.50SampleOutput123005AuthorlcySourceC语言程序设计练习〔二RecommendJGShining解答：#include<stdio.h>intmain<>{intn,i,b1,b2,b3;doublea[101];while<scanf<"%d",&n>!=EOF&&n!=0>{for<i=0;i<n;i++>scanf<"%lf",&a[i]>;b1=b2=b3=0;for<i=0;i<n;i++>{if<a[i]<0>b1++;elseif<a[i]==0>b2++;elseb3++;}printf<"%d%d%d\n",b1,b2,b3>;}}2009求数列的和ProblemDescription数列的定义如下：

数列的第一项为n,以后各项为前一项的平方根,求数列的前m项的和。Input输入数据有多组,每组占一行,由两个整数n〔n<10000和m<m<1000>组成,n和m的含义如前所述。Output对于每组输入数据,输出该数列的和,每个测试实例占一行,要求精度保留2位小数。SampleInput81422SampleOutput94.733.41AuthorlcySourceC语言程序设计练习〔二RecommendJGShining解答：#include<stdio.h>#include<math.h>main<>{doublen,m,s,w,i;while<scanf<"%lf%lf",&n,&m>!=EOF>{s=n;for<i=1;i<m;i++>{n=sqrt<n>;s=s+n;}printf<"%.2lf\n",s>;}}2010水仙花数ProblemDescription春天是鲜花的季节,水仙花就是其中最迷人的代表,数学上有个水仙花数,他是这样定义的：

"水仙花数"是指一个三位数,它的各位数字的立方和等于其本身,比如：153=1^3+5^3+3^3。

现在要求输出所有在m和n围的水仙花数。Input输入数据有多组,每组占一行,包括两个整数m和n〔100<=m<=n<=999。Output对于每个测试实例,要求输出所有在给定围的水仙花数,就是说,输出的水仙花数必须大于等于m,并且小于等于n,如果有多个,则要求从小到大排列在一行输出,之间用一个空格隔开;

如果给定的围不存在水仙花数,则输出no;

每个测试实例的输出占一行。SampleInput100120300380SampleOutputno370371AuthorlcySourceC语言程序设计练习〔二RecommendJGShining解答：#include<stdio.h>main<>{intm,n,i,w,a,b,c,j,s,d;while<scanf<"%d%d",&n,&m>!=EOF>{d=0;j=1;if<m>n>{w=m;m=n;n=w;}else;for<i=m;i<=n;i++>{a=i/100;b=i/10%10;c=i%10;s=a*a*a+b*b*b+c*c*c;if<i==s>{if<d!=0>printf<"">;printf<"%d",i>;d=d+1;j=j+1;}}if<j==1>printf<"no\n">;elseprintf<"\n">;}}2011多项式求和ProblemDescription多项式的描述如下：

1-1/2+1/3-1/4+1/5-1/6+...

现在请你求出该多项式的前n项的和。Input输入数据由2行组成,首先是一个正整数m〔m<100,表示测试实例的个数,第二行包含m个正整数,对于每一个整数<不妨设为n,n<1000,求该多项式的前n项的和。Output对于每个测试实例n,要求输出多项式前n项的和。每个测试实例的输出占一行,结果保留2位小数。SampleInput212SampleOutput1.000.50AuthorlcySourceC语言程序设计练习〔二RecommendJGShining解答：#include<stdio.h>#include<math.h>main<>{doublem,n,i,s,j,k,a;while<scanf<"%lf",&m>!=EOF>{for<i=0;i<m;i++>{s=0;scanf<"%lf",&n>;for<j=1;j<=n;j++>s=s+1/j*pow<-1,j+1>;printf<"%.2lf\n",s>;}}}2012素数判定ProblemDescription对于表达式n^2+n+41,当n在〔x,y围取整数值时〔包括x,y<-39<=x<y<=50>,判定该表达式的值是否都为素数。Input输入数据有多组,每组占一行,由两个整数x,y组成,当x=0,y=0时,表示输入结束,该行不做处理。Output对于每个给定围的取值,如果表达式的值都为素数,则输出"OK",否则请输出"Sorry",每组输出占一行。SampleInput0100SampleOutputOKAuthorlcySourceC语言程序设计练习〔二RecommendJGShining解答：#include<stdio.h>main<>{intx,y,i,j,s,k,w,d;while<scanf<"%d%d",&x,&y>==2&&<x!=0||y!=0>>{w=0;for<i=x;i<=y;i++>{k=i*i+i+41;for<j=2;j<k;j++>{d=k%j;if<d==0>w++;}}if<w==0>printf<"OK\n">;elseprintf<"Sorry\n">;}}2014青年歌手大奖赛_评委会打分ProblemDescription青年歌手大奖赛中,评委会给参赛选手打分。选手得分规则为去掉一个最高分和一个最低分,然后计算平均得分,请编程输出某选手的得分。Input输入数据有多组,每组占一行,每行的第一个数是n<2<n<100>,表示评委的人数,然后是n个评委的打分。Output对于每组输入数据,输出选手的得分,结果保留2位小数,每组输出占一行。SampleInput39998974100999897SampleOutput98.0098.50AuthorlcySourceC语言程序设计练习〔三Recommendlcy解答：#include<stdio.h>intmain<>{intn,s,a[100],i,k,b;doublew;while<scanf<"%d",&n>!=EOF>{k=0;w=0;s=0;for<i=0;i<n;i++>{scanf<"%d",&a[i]>;k++;b=a[0];w=w+a[i];}for<i=0;i<k;i++>{if<a[i]>s>s=a[i];}for<i=1;i<k;i++>{if<b>a[i]>b=a[i];}w=<w-s-b>/<k-2>;printf<"%.2lf\n",w>;}}2015偶数求和ProblemDescription有一个长度为n<n<=100>的数列,该数列定义为从2开始的递增有序偶数,现在要求你按照顺序每m个数求出一个平均值,如果最后不足m个,则以实际数量求平均值。编程输出该平均值序列。Input输入数据有多组,每组占一行,包含两个正整数n和m,n和m的含义如上所述。Output对于每组输入数据,输出一个平均值序列,每组输出占一行。SampleInput3242SampleOutput3637AuthorlcySourceC语言程序设计练习〔三Recommendlcy解答：#include<stdio.h>main<>{intn,m,a,b,i,j,k,w,l,e,s,d,r;while<scanf<"%d%d",&n,&m>!=EOF>{s=0;e=0;l=0;if<n<=m>{for<i=0;i<n;i++>{s=s+2;e=e+s;k=e/n;}printf<"%d\n",k>;}else{w=n%m;r=0;for<i=1;i<=n-w;i++>{s=s+2;l=l+s;e=e+s;if<i%m==0>{k=e/m;e=0;if<r>printf<"">;printf<"%d",k>;r=r+1;}}s=0;if<w!=0>{for<j=0;j<n;j++>{s=s+2;e=e+s;}d=e-l;k=d/w;printf<"">;printf<"%d",k>;}printf<"\n">;}}}2016数据的交换输出ProblemDescription输入n<n<100>个数,找出其中最小的数,将它与最前面的数交换后输出这些数。Input输入数据有多组,每组占一行,每行的开始是一个整数n,表示这个测试实例的数值的个数,跟着就是n个整数。n=0表示输入的结束,不做处理。Output对于每组输入数据,输出交换后的数列,每组输出占一行。SampleInput421345543210SampleOutput123414325AuthorlcySourceC语言程序设计练习〔三Recommendlcy解答：#include<stdio.h>main<>{intn,a[100],i,j,k,s,w;while<scanf<"%d",&n>!=EOF&&n!=0>{j=0;for<i=0;i<n;i++>{scanf<"%d",&a[i]>;k=a[0];}for<i=0;i<n;i++>{if<k>a[i]>{k=a[i];j=i;}}w=a[0];a[0]=k;a[j]=w;for<i=0;i<n;i++>{printf<"%d",a[i]>;if<n-i!=1>printf<"">;}printf<"\n">;}}2017字符串统计ProblemDescription对于给定的一个字符串,统计其中数字字符出现的次数。Input输入数据有多行,第一行是一个整数n,表示测试实例的个数,后面跟着n行,每行包括一个由字母和数字组成的字符串。Output对于每个测试实例,输出该串中数值的个数,每个输出占一行。SampleInput2asdfasdf123123asdfasdfasdf111111111asdfasdfasdfSampleOutput69Authorlcy解答：#include<stdio.h>main<>{intn,i,j,a;chars[1000];while<scanf<"%d",&n>!=EOF>{getchar<>;for<i=0;i<n;i++>{gets<s>;a=0;for<j=0;s[j]!='\0';j++>{if<<s[j]<='9'>&&<s[j]>='0'>>a=a+1;}printf<"%d\n",a>;}}}2019数列有序!ProblemDescription有n<n<=100>个整数,已经按照从小到大顺序排列好,现在另外给一个整数x,请将该数插入到序列中,并使新的序列仍然有序。Input输入数据包含多个测试实例,每组数据由两行组成,第一行是n和m,第二行是已经有序的n个数的数列。n和m同时为0标示输入数据的结束,本行不做处理。Output对于每个测试实例,输出插入新的元素后的数列。SampleInput3312400SampleOutput1234AuthorlcySourceC语言程序设计练习〔三Recommendlcy解答：#include<stdio.h>main<>{intn,m,a[100],b[100],i,j,k,s,w,d;scanf<"%d%d",&n,&m>;while<!<n==0&&m==0>>{w=0;for<i=0;i<n;i++>scanf<"%d",&a[i]>;s=a[0];if<m<s>{printf<"%d",m>;for<j=0;j<n;j++>{printf<"">;printf<"%d",a[j]>;}}else{for<j=0;j<n;j++>{if<m>a[j]>w=w+1;}for<j=0;j<w;j++>{printf<"%d",a[j]>;printf<"">;}printf<"%d",m>;for<j=w;j<n;j++>{printf<"">;printf<"%d",a[j]>;}}printf<"\n">;scanf<"%d%d",&n,&m>;}}2020绝对值排序ProblemDescription输入n<n<=100>个整数,按照绝对值从大到小排序后输出。题目保证对于每一个测试实例,所有的数的绝对值都不相等。Input输入数据有多组,每组占一行,每行的第一个数字为n,接着是n个整数,n=0表示输入数据的结束,不做处理。Output对于每个测试实例,输出排序后的结果,两个数之间用一个空格隔开。每个测试实例占一行。SampleInput33-424012-30SampleOutput-432-3210AuthorlcySourceC语言程序设计练习〔三Recommendlcy解答：#include<stdio.h>main<>{intn,m,a[100],b[100],c,d,e,f,i,j,k;while<scanf<"%d",&n>!=EOF&&n!=0>{for<i=0;i<n;i++>scanf<"%d",&a[i]>;f=0;for<j=0;j<n;j++>{c=0;for<i=0;i<n;i++>{if<a[i]<0>m=-a[i];elsem=a[i];if<c<=m>{c=m;b[j]=a[i];k=i;}}a[k]=0;if<f>printf<"">;printf<"%d",b[j]>;f=f+1;}printf<"\n">;}}2021发工资咯：ProblemDescription作为杭电的老师,最盼望的日子就是每月的8号了,因为这一天是发工资的日子,养家糊口就靠它了,呵呵

但是对于学校财务处的工作人员来说,这一天则是很忙碌的一天,财务处的小胡老师最近就在考虑一个问题：如果每个老师的工资额都知道,最少需要准备多少人民币,才能在给每位老师发工资的时候都不用老师找零呢？

这里假设老师的工资都是正整数,单位元,人民币一共有100元、50元、10元、5元、2元和1元六种。Input输入数据包含多个测试实例,每个测试实例的第一行是一个整数n〔n<100,表示老师的人数,然后是n个老师的工资。

n=0表示输入的结束,不做处理。Output对于每个测试实例输出一个整数x,表示至少需要准备的人民币数。每个输出占一行。SampleInput31230SampleOutput4AuthorlcySourceC语言程序设计练习〔四Recommendlcy解答：#include<stdio.h>main<>{intn,m,a,b,c,d,e,f,i,j,k;while<scanf<"%d",&n>!=EOF&&n!=0>{k=0;for<i=0;i<n;i++>{scanf<"%d",&m