杭电acm答案无敌版(共54页)

1、精选优质文档-倾情为你奉上 选修课考试作业姓名：春杰级：电商14专心-专注-专业1001 Sum ProblemProblem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + . + n.InputThe input will consist of a series of integers n, one integer per line.OutputFor each ca

2、se, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.Sample Input1100Sample Output15050AuthorDOOM III解答：#includemain() int n,i,sum; sum=0; while(scanf(%d,&n)!=-1) sum=0; for(i=0;i=n;i+) sum+=i; printf(%dnn,sum); 1089 A+B for

3、 Input-Output Practice (I)Problem DescriptionYour task is to Calculate a + b.Too easy?! Of course! I specially designed the problem for acm beginners. You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim. InputThe input wil

4、l consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line. OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input. Sample Input1 510 20Sample Output630AuthorlcyRe

5、commendJGShining解答：#include main() int a,b; while(scanf(%d%d,&a,&b)!=EOF) printf(%dn,a+b); 1090 A+B for Input-Output Practice (II)Problem DescriptionYour task is to Calculate a + b.InputInput contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a

6、 and b, separated by a space, one pair of integers per line. OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input. Sample Input21 510 20Sample Output630AuthorlcyRecommendJGShining 解答：#include#define M 1000v

7、oid main() int a ,b,n,jM,i; /printf(please input n:n); scanf(%d,&n); for(i=0;in;i+) scanf(%d%d,&a,&b); /printf(%d %d,a,b); ji=a+b; i=0; while(in) printf(%d,ji); i+; printf(n); 1091 A+B for Input-Output Practice (III)Problem DescriptionYour task is to Calculate a + b.InputInput contains multiple test

8、 cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of ou

9、tput for each line in input. Sample Input1 510 200 0Sample Output630AuthorlcyRecommendJGShining解答：#include main() int a,b; scanf(%d %d,&a,&b); while(!(a=0&b=0) printf(%dn,a+b); scanf(%d %d,&a,&b); 1092 A+B for Input-Output Practice (IV)Problem DescriptionYour task is to Calculate the sum of some int

10、egers.InputInput contains multiple test cases. Each test case contains a integer N, and then N integers follow in the same line. A test case starting with 0 terminates the input and this test case is not to be processed.OutputFor each group of input integers you should output their sum in one line,

11、and with one line of output for each line in input. Sample Input4 1 2 3 45 1 2 3 4 50 Sample Output1015AuthorlcyRecommendJGShining解答：#include int main() int n,sum,i,t; while(scanf(%d,&n)!=EOF&n!=0) sum=0; for(i=0;in;i+) scanf(%d,&t); sum=sum+t; printf(%dn,sum); 1093 A+B for Input-Output Practice (V)

12、Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line. OutputFor each group of input integers you should output their sum in one

13、 line, and with one line of output for each line in input. Sample Input24 1 2 3 45 1 2 3 4 5Sample Output1015Authorlcy解答：#includemain() int n,a,b,i,j,sum; sum=0; while(scanf(%dn,&n)!=-1) for(i=0;in;i+) scanf(%d,&b); for(j=0;jb;j+) scanf(%d,&a); sum+=a; printf(%dn,sum); sum=0; 1094 A+B for Input-Outp

14、ut Practice (VI)Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line. OutputFor each test case you should output the sum of N integers in o

15、ne line, and with one line of output for each line in input. Sample Input4 1 2 3 45 1 2 3 4 5Sample Output1015AuthorlcyRecommendJGShining解答：#includemain() int n,a,b,i,j,sum; sum=0; while(scanf(%dn,&n)!=-1) for(j=0;jn;j+) scanf(%d,&a); sum+=a; printf(%dn,sum); sum=0; 1095 A+B for Input-Output Practic

16、e (VII)Problem DescriptionYour task is to Calculate a + b.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line. OutputFor each pair of input integers a and b you should output the sum of a and b, and followed by a blank line. Sampl

17、e Input1 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include main() int a,b; while(scanf(%d%d,&a,&b)!=EOF) printf(%dnn,a+b); 1096 A+B for Input-Output Practice (VIII)Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and

18、 then N lines follow. Each line starts with a integer M, and then M integers follow in the same line. OutputFor each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.Sample Input34 1 2 3 45 1 2 3 4 53 1 2 3Sample Output1015

19、6AuthorlcyRecommendJGShining解答：int main() int a,b,i,j,l1000,k; scanf(%d,&i); getchar(); for(j=1;j=i;j+) lj=0; for(j=1;j=i;j+) scanf(%d,&a); getchar(); for(k=1;k=a;k+) scanf(%d,&b); getchar(); lj+=b; for(j=1;j=i-1;j+) printf(%dnn,lj); printf(%dn,li); 2000 ASCII码排序Problem Description输入三个字符后，按各字符的ASCII

20、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。Sample InputqweasdzxcSample Outpute q wa d sc x zAuthorlcySourceRecommendJGShining解答：#includemain() char a,b,c,d; while(scanf(%c %c %c,&a,&b,&c)!=EOF) getchar(); if(a=b) if(c=a) printf(%c %c %cn,b,a,c); else if(b=c) pri

21、ntf(%c %c %cn,c,b,a); else if(b=b) printf(%c %c %cn,a,b,c); else if(c=a) printf(%c %c %cn,a,c,b); else if(ac) printf(%c %c %cn,c,a,b); 2001计算两点间的距离Problem Description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。Sample Input0 0 0

22、10 1 1 0Sample Output1.001.41AuthorlcySourceRecommendJGShining解答：#include#includemain() double a,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(%.2lfn,s); 2002计算球体积Problem Description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

23、组输入数据，输出一行，计算结果保留三位小数。Sample Input11.5Sample Output4.18914.137Hint#define PI 3. AuthorlcySourceRecommendJGShining解答：#include#define PI 3.main() double a,v; while(scanf(%lf,&a)!=EOF) v=4*PI*a*a*a/3; printf(%.3lfn,v); 2003求绝对值Problem Description求实数的绝对值。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

24、值，要求每组数据输出一行，结果保留两位小数。Sample Input123-234.00Sample Output123.00234.00AuthorlcySourceRecommendJGShining解答：#includemain() double a; while(scanf(%lf,&a)!=EOF) if(a0) a=-a; printf(%.2lfn,a); 2004成绩转换Problem Description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：90100为A;8089为B;7079为C;6069为D;059为E;Input输入数据有多组，每组占一行，

25、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0100范围内，请输出一行：“Score is error!”。Sample Input5667100123Sample OutputEDAScore is error!AuthorlcySourceRecommendJGShining解答：#include int main() int n; while(scanf(%d,&n)!=EOF) if(n100|n=90)printf(An); else if(n=80)printf(Bn); else if(n=70)printf(Cn); else if(n=60)pri

26、ntf(Dn); else printf(En); return 0;2005第几天？Problem Description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sample input ,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。Sample Input1985/1/202006/3/12Sample Output2071AuthorlcySourceRecommendJGShining解答：#includemain() int a,b,c,

27、d,e,f,g; while(scanf(%d/%d/%d,&a,&b,&c)!=EOF) if(b=1) d=c; else if(b=2) d=31+c; else if(b=3) d=31+28+c; else if(b=4) d=31+28+31+c; else if(b=5) d=31+31+28+30+c; else if(b=6) d=31+28+31+30+31+c; else if(b=7) d=31+28+31+30+31+30+c; else if(b=8) d=31+28+31+30+31+30+31+c; else if(b=9) d=31+28+31+30+31+3

28、0+31+31+c; else if(b=10) d=31+28+31+30+31+30+31+31+30+c; else if(b=11) d=31+28+31+30+31+30+31+31+30+31+c; else if(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; else d=d; else if(g=0) d=d+1; else d=d; printf(%dn,d); 2006求奇数的乘积Problem Description给你n个整数，求他们

29、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。Sample Input3 1 2 34 2 3 4 5Sample Output315AuthorlcySourceRecommendJGShining解答：#includemain() int n,s,i,a; while(scanf(%d,&n)!=EOF) s=1; for(i=0;in;i+) scanf(%d,&a); if(a%2=1) s=s

30、*a; else ; printf(%dn,s); 2007平方和与立方和Problem Description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。Sample Input1 32 5Sample Output4 2820 152AuthorlcySourceRecommendJGShining解答：#includeint ma

31、in() int sum1,sum2,n,i,m,t; while(scanf(%d%d,&m,&n)!=EOF) sum1=sum2=0; if(mn)t=m;m=n;n=t; for(i=m;i=n;i+) if(i%2=0) sum1+=(i*i); else sum2+=(i*i*i); printf(%d %dn,sum1,sum2); return 0;2008数值统计Problem Description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（n100），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

32、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。Sample Input6 0 1 2 3 -1 05 1 2 3 4 0.50 Sample Output1 2 30 0 5AuthorlcySourceRecommendJGShining解答：#includeint main() int n,i,b1,b2,b3; double a101; while(scanf(%d,&n)!=EOF & n!=0) for(i=0;in;i+) scanf(%lf,&ai); b1=b2=b3=0; for(i=0;in;i+) if(

33、ai0) b1+; else if(ai=0) b2+; else b3+; printf(%d %d %dn,b1,b2,b3); 2009求数列的和Problem Description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（n10000）和m(m1000)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。Sample Input81 42 2Sample Output94.733.41AuthorlcySourceRecommend

34、JGShining解答：#include#includemain() double n,m,s,w,i; while(scanf(%lf%lf,&n,&m)!=EOF) s=n; for(i=1;im;i+) n=sqrt(n); s=s+n; printf(%.2lfn,s); 2010水仙花数Problem Description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=13+53+33。现在要求输出所有在m和n范围内的水仙花数。Input输入数据有多组，每组占一行，包括两个整

35、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范围内的水仙花数，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内输出，之间用一个空格隔开;如果给定的范围内不存在水仙花数，则输出no;每个测试实例的输出占一行。Sample Input100 120300 380Sample Outputno370 371AuthorlcySourceRecommendJGShining解答：#includemain() int m,n,i,w,a,b,c,j,s,d; while(scanf(%d %d,&n,&m)!=EOF)

36、d=0; j=1; if(mn) 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(non); else printf(n); 2011多项式求和Problem Description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

37、，首先是一个正整数m（m100），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,n1000），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。Sample Input21 2Sample Output1.000.50AuthorlcySourceRecommendJGShining解答：#include#includemain() double m,n,i,s,j,k,a; while(scanf(%lf,&m)!=EOF) for(i=0;im;i+) s=0; scanf(%lf,&n);

38、 for(j=1;j=n;j+) s=s+1/j*pow(-1,j+1); printf(%.2lfn,s); 2012素数判定Problem Description对于表达式n2+n+41，当n在（x,y）范围内取整数值时（包括x,y）(-39=xy=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对于每个给定范围内的取值，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。Sample Input0 10 0Sample OutputOKAuthorlcy

39、SourceRecommendJGShining解答：#includemain() int x,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;jk;j+) d=k%j; if(d=0) w+; if(w=0) printf(OKn); else printf(Sorryn); 2014青年歌手大奖赛_评委会打分Problem Description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(2n100)，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。