杭電ACM部分題目答案

1、 杭電ACM1001 Sum Problem21089 A+B for Input-Output Practice (I)41090 A+B for Input-Output Practice (II)61091 A+B for Input-Output Practice (III)81092 A+B for Input-Output Practice (IV)91093 A+B for Input-Output Practice (V)111094 A+B for Input-Output Practice (VI)121095 A+B for Input-Output Practice (

2、VII)131096 A+B for Input-Output Practice (VIII)142000 ASCII碼排序162001計(jì)算兩點(diǎn)間的距離172002計(jì)算球體積192003求絕對(duì)值202004成績(jī)轉(zhuǎn)換212005第幾天？222006求奇數(shù)的乘積242007平方和與立方和262008數(shù)值統(tǒng)計(jì)272009求數(shù)列的和282010水仙花數(shù)292011多項(xiàng)式求和312012素?cái)?shù)判定332014青年歌手大獎(jiǎng)賽_評(píng)委會(huì)打分342015偶數(shù)求和362016數(shù)據(jù)的交換輸出382017字符串統(tǒng)計(jì)402019數(shù)列有序!412020絕對(duì)值排序432021發(fā)工資咯：）452033人見(jiàn)人愛(ài)A+B462039

3、三角形482040親和數(shù)49姓名：鄭春杰班級(jí)：電商1001學(xué)號(hào)：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 case, outp

4、ut 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 Input-O

5、utput 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 consis

6、t 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 Output630AuthorlcyRecommendJ

7、GShining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 and b,

8、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 main

9、() 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 cases.

10、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 output for

11、 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 integers.In

12、putInput 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, and with

13、 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)Problem

14、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 line, a

15、nd 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-Output Pract

16、ice (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 one line,

17、 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; Copy to Clipboard Save to File1095 A+B fo

18、r 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 integers a and b you should output the sum of a and b, and followed

19、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 calculate the sum of some integers.InputInput contains an integer N

20、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 is a blank line between outputs.Sample Input34 1 2 3 45 1 2 3 4 53

21、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;j+) printf(%dnn,lj); printf(%dn,li); 2000 ASCII碼排序Problem Descrip

22、tion輸入三個(gè)字符后，按各字符的ASCII碼從小到大的順序輸出這三個(gè)字符。Input輸入數(shù)據(jù)有多組，每組占一行，有三個(gè)字符組成，之間無(wú)空格。Output對(duì)于每組輸入數(shù)據(jù)，輸出一行，字符中間用一個(gè)空格分開(kāi)。Sample InputqweasdzxcSample Outpute q wa d sc x zAuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） RecommendJGShining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

23、(%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計(jì)算兩點(diǎn)間的距離Problem Description輸入兩點(diǎn)坐標(biāo)（X1,Y1）,（X2,Y2）,計(jì)算并輸出兩點(diǎn)間的距離。Input輸入數(shù)據(jù)有多組，每組占一行，由4個(gè)實(shí)數(shù)組成，分別表示x1,y1,x2,y2,數(shù)據(jù)之間用空格隔開(kāi)。Output對(duì)于每組輸入

24、數(shù)據(jù)，輸出一行，結(jié)果保留兩位小數(shù)。Sample Input0 0 0 10 1 1 0Sample Output1.001.41AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（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(%.2lfn,s); 2002計(jì)算球體積Problem Description根據(jù)輸入的半徑值，計(jì)算球的體積。Input輸

25、入數(shù)據(jù)有多組，每組占一行，每行包括一個(gè)實(shí)數(shù)，表示球的半徑。Output輸出對(duì)應(yīng)的球的體積，對(duì)于每組輸入數(shù)據(jù)，輸出一行，計(jì)算結(jié)果保留三位小數(shù)。Sample Input11.5Sample Output4.18914.137Hint#define PI 3. AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） RecommendJGShining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求絕對(duì)值Problem Descrip

26、tion求實(shí)數(shù)的絕對(duì)值。Input輸入數(shù)據(jù)有多組，每組占一行，每行包含一個(gè)實(shí)數(shù)。Output對(duì)于每組輸入數(shù)據(jù)，輸出它的絕對(duì)值，要求每組數(shù)據(jù)輸出一行，結(jié)果保留兩位小數(shù)。Sample Input123-234.00Sample Output123.00234.00AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） RecommendJGShining解答：#includemain() double a; while(scanf(%lf,&a)!=EOF) if(a0) a=-a; printf(%.2lfn,a); 2004成績(jī)轉(zhuǎn)換Problem Description輸入一個(gè)百分制的成績(jī)t，

27、將其轉(zhuǎn)換成對(duì)應(yīng)的等級(jí)，具體轉(zhuǎn)換規(guī)則如下：90100為A;8089為B;7079為C;6069為D;059為E;Input輸入數(shù)據(jù)有多組，每組占一行，由一個(gè)整數(shù)組成。Output對(duì)于每組輸入數(shù)據(jù)，輸出一行。如果輸入數(shù)據(jù)不在0100范圍內(nèi)，請(qǐng)輸出一行：“Score is error!”。Sample Input5667100123Sample OutputEDAScore is error!AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） RecommendJGShining解答：#include int main() int n; while(scanf(%d,&n)!=EOF) if(n

28、100|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第幾天？Problem Description給定一個(gè)日期，輸出這個(gè)日期是該年的第幾天。Input輸入數(shù)據(jù)有多組，每組占一行，數(shù)據(jù)格式為YYYY/MM/DD組成，具體參見(jiàn)sample input ,另外，可以向你確保所有的輸入數(shù)據(jù)是合法的。Output對(duì)于每組輸入數(shù)據(jù)，輸出一行，表示該日期是該年的第幾天。Sample Input1985/1/20

29、2006/3/12Sample Output2071AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（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+31+30+31+c; else i

30、f(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

31、=1+d; else d=d; else if(g=0) d=d+1; else d=d; printf(%dn,d); 2006求奇數(shù)的乘積Problem Description給你n個(gè)整數(shù)，求他們中所有奇數(shù)的乘積。Input輸入數(shù)據(jù)包含多個(gè)測(cè)試實(shí)例，每個(gè)測(cè)試實(shí)例占一行，每行的第一個(gè)數(shù)為n，表示本組數(shù)據(jù)一共有n個(gè)，接著是n個(gè)整數(shù)，你可以假設(shè)每組數(shù)據(jù)必定至少存在一個(gè)奇數(shù)。Output輸出每組數(shù)中的所有奇數(shù)的乘積，對(duì)于測(cè)試實(shí)例，輸出一行。Sample Input3 1 2 34 2 3 4 5Sample Output315AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） Recommen

32、dJGShining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*a; else ; printf(%dn,s); 2007平方和與立方和Problem Description給定一段連續(xù)的整數(shù)，求出他們中所有偶數(shù)的平方和以及所有奇數(shù)的立方和。Input輸入數(shù)據(jù)包含多組測(cè)試實(shí)例，每組測(cè)試實(shí)例包含一行，由兩個(gè)整數(shù)m和n組成。Output對(duì)于每組輸入數(shù)據(jù)，輸出一行，應(yīng)包括兩個(gè)整數(shù)x和y，分別表示該段連續(xù)的整數(shù)中所有偶數(shù)的平方和以及所有奇數(shù)

33、的立方和。你可以認(rèn)為32位整數(shù)足以保存結(jié)果。Sample Input1 32 5Sample Output4 2820 152AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（一） RecommendJGShining解答：#includeint main() 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);

34、 return 0;2008數(shù)值統(tǒng)計(jì)Problem Description統(tǒng)計(jì)給定的n個(gè)數(shù)中，負(fù)數(shù)、零和正數(shù)的個(gè)數(shù)。Input輸入數(shù)據(jù)有多組，每組占一行，每行的第一個(gè)數(shù)是整數(shù)n（n100），表示需要統(tǒng)計(jì)的數(shù)值的個(gè)數(shù)，然后是n個(gè)實(shí)數(shù)；如果n=0，則表示輸入結(jié)束，該行不做處理。Output對(duì)于每組輸入數(shù)據(jù)，輸出一行a,b和c，分別表示給定的數(shù)據(jù)中負(fù)數(shù)、零和正數(shù)的個(gè)數(shù)。Sample Input6 0 1 2 3 -1 05 1 2 3 4 0.50 Sample Output1 2 30 0 5AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（二） RecommendJGShining解答：#inc

35、ludeint 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(ai0) b1+; else if(ai=0) b2+; else b3+; printf(%d %d %dn,b1,b2,b3); 2009求數(shù)列的和Problem Description數(shù)列的定義如下：數(shù)列的第一項(xiàng)為n，以后各項(xiàng)為前一項(xiàng)的平方根，求數(shù)列的前m項(xiàng)的和。Input輸入數(shù)據(jù)有多組，每組占一行，由兩個(gè)整數(shù)n（

36、n10000）和m(m1000)組成，n和m的含義如前所述。Output對(duì)于每組輸入數(shù)據(jù)，輸出該數(shù)列的和，每個(gè)測(cè)試實(shí)例占一行，要求精度保留2位小數(shù)。Sample Input81 42 2Sample Output94.733.41AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（二） RecommendJGShining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

37、數(shù)Problem Description春天是鮮花的季節(jié)，水仙花就是其中最迷人的代表，數(shù)學(xué)上有個(gè)水仙花數(shù)，他是這樣定義的：“水仙花數(shù)”是指一個(gè)三位數(shù)，它的各位數(shù)字的立方和等于其本身，比如：153=13+53+33。現(xiàn)在要求輸出所有在m和n范圍內(nèi)的水仙花數(shù)。Input輸入數(shù)據(jù)有多組，每組占一行，包括兩個(gè)整數(shù)m和n（100=m=n=999）。Output對(duì)于每個(gè)測(cè)試實(shí)例，要求輸出所有在給定范圍內(nèi)的水仙花數(shù)，就是說(shuō)，輸出的水仙花數(shù)必須大于等于m,并且小于等于n，如果有多個(gè)，則要求從小到大排列在一行內(nèi)輸出，之間用一個(gè)空格隔開(kāi);如果給定的范圍內(nèi)不存在水仙花數(shù)，則輸出no;每個(gè)測(cè)試實(shí)例的輸出占一行。Sam

38、ple Input100 120300 380Sample Outputno370 371AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（二） RecommendJGShining解答：#includemain() int m,n,i,w,a,b,c,j,s,d; while(scanf(%d %d,&n,&m)!=EOF) 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( ); pri

39、ntf(%d,i); d=d+1; j=j+1; if(j=1) printf(non); else printf(n); 2011多項(xiàng)式求和Problem Description多項(xiàng)式的描述如下：1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + .現(xiàn)在請(qǐng)你求出該多項(xiàng)式的前n項(xiàng)的和。Input輸入數(shù)據(jù)由2行組成，首先是一個(gè)正整數(shù)m（m100），表示測(cè)試實(shí)例的個(gè)數(shù)，第二行包含m個(gè)正整數(shù)，對(duì)于每一個(gè)整數(shù)(不妨設(shè)為n,n1000），求該多項(xiàng)式的前n項(xiàng)的和。Output對(duì)于每個(gè)測(cè)試實(shí)例n，要求輸出多項(xiàng)式前n項(xiàng)的和。每個(gè)測(cè)試實(shí)例的輸出占一行，結(jié)果保留2位小數(shù)。Sample Input2

40、1 2Sample Output1.000.50AuthorlcySourceC語(yǔ)言程序設(shè)計(jì)練習(xí)（二） RecommendJGShining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); for(j=1;j=n;j+) s=s+1/j*pow(-1,j+1); printf(%.2lfn,s); 2012素?cái)?shù)判定Problem Description對(duì)于表達(dá)式n2+n+41，當(dāng)n在（x,y）范圍內(nèi)取整數(shù)值時(shí)（包括x,y）(-39=xy=50)，判定該表達(dá)式的值是否都為素?cái)?shù)。Input輸入數(shù)據(jù)有多組，每組占一行，由兩個(gè)整數(shù)x，y組成，當(dāng)x=0,y=0時(shí)，表示輸入結(jié)束，該行不做處理。Output對(duì)于每個(gè)給定范圍內(nèi)的取值，如果表達(dá)式的值都為素?cái)?shù)，則輸出OK,否則請(qǐng)輸出“Sorry”,