杭电题目acm答案.docx

1、杭电题目acm答案 选修课考试作业:姓名：郑春杰班级：电商1001学号：341001 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 case

2、, 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 I

3、nput-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 will c

4、onsist 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 Output630AuthorlcyRecom

5、mendJGShining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 an

6、d 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 1000void

7、 main() int a ,b,n,jM,i; InputInput contains multiple test 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 outp

8、ut the sum of a and b in one line, and with one line of output 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)Proble

9、m DescriptionYour task is to Calculate the sum of some integers.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 o

10、f input integers you should output their sum in one line, 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+

11、t; printf(%dn,sum); 1093 A+B for Input-Output Practice (V)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

12、group of input integers you should output their sum in one 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

13、); sum+=a; printf(%dn,sum); sum=0; 1094 A+B for Input-Output 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 e

14、ach test case you should output the sum of N integers in one 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;

15、 printf(%dn,sum); sum=0; Copy to Clipboard Save to File1095 A+B for Input-Output Practice (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 in

16、tegers a and b you should output the sum of a and b, and followed by a blank line. Sample 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 ca

17、lculate 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 line, and you must note that there i

18、s a blank line between outputs.Sample Input34 1 2 3 45 1 2 3 4 53 1 2 3Sample Output10156AuthorlcyRecommendJGShining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

19、;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 zAuthorlcySourceC语言程序设计练习（一） RecommendJGShining解答：#includemain() char a,b,c,d; whil

20、e(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) printf(%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

21、。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 10 1 1 0Sample OutputAuthorlcySourceC语言程序设计练习（一） RecommendJGShining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(%.2

22、lfn,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，对于每组输入数据，输出一行，计算结果保留三位小数。Sample Input1Sample OutputHint#define PI AuthorlcySourceC语言程序设计练习（一） RecommendJGShining解答：#include#define PI main() double a,v; while(scanf(%lf,&a)!=EOF) v=4*PI*a*a*a/3; pr

23、intf(%.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，输出它的绝对值，要求每组数据输出一行，结果保留两位小数。Sample Input123Sample OutputAuthorlcySourceC语言程序设计练习（一） RecommendJGShining解答：#includemain() double a; while(scanf(%lf,&a)!=EOF) if(a0) a=-a; printf(%.2lfn,a); 2004成绩转换Problem Des

24、cription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，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!AuthorlcySourceC语言程序设计练习（一） RecommendJGShining解答：#include int main() int n; while(sca

25、nf(%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)printf(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。Sa

26、mple Input1985/1/202006/3/12Sample Output2071AuthorlcySourceC语言程序设计练习（一） RecommendJGShining解答：#includemain() int a,b,c,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

27、+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+30+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=