杭电OJ的输入输出格式题.docx
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杭电OJ的输入输出格式题
杭电OJ的输入/输出格式题
输入
一、1000A+BProblem
题目名称:
A+BProblem
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
TimeLimit:
2000/1000MS(Java/Others) MemoryLimit:
65536/32768K(Java/Others)
ProblemDescription
CalculateA+B.
Input
EachlinewillcontaintwointegersAandB.Processtoendoffile.
Output
Foreachcase,outputA+Binoneline.
SampleInput
11
SampleOutput
2
参考答案
#include
intmain(void)
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF)
printf("%d\n",a+b);
return0;
}
二、1002A+BProblemII
题目名称:
A+BProblemII
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
Time Limit:
2000/1000 MS (Java/Others) Memory Limit:
65536/32768 K (Java/Others)
Total Submission(s):
22627 Accepted Submission(s):
4035
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:
", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 111111*********1110
思路:
本题是手工模拟两个大数相加的问题。
两个输入的数字用字符数组存储。
不超过1000位,也就是说用整型数定义是不行的,然后想到的就是用整型数组存储和的各位数。
具体解题思路如下:
1、先定义两个字符型数组str1和str2,用以保存输入的数字,然后再定义一个整型数组,保存两加数的和的各位。
2、下面来求和计算,先将字符型数组str1中的各位从后面倒着取出,转换成数字存储在整型数组a中(即按相反的顺序进行存放),然后,再将字符型数组str2中的各位从后面倒着取出,转换成数字,然后与a数组中的各位相加,并加入进位,这样相当于将两个数从最低位到最高位进行相加,但是数组a中存放的和是从最低位到最高位,所以输出时要注意。
3、最后就是格式控制好输出。
参考答案
#include
#include
#include
intmain()
{
charstr1[1005],str2[1005];
intn,count=0,i,j,flag;
inta[1005];
scanf("%d",&n);
while(n--)
{
scanf("%s%s",str1,str2);
memset(a,0,sizeof(a));
for(i=strlen(str1)-1,j=0;i>=0;i--,j++)
a[j]=str1[i]-'0';
for(i=strlen(str2)-1,j=0;i>=0;i--,j++)
{
a[j]=a[j]+str2[i]-'0';
a[j+1]=a[j+1]+a[j]/10;
a[j]=a[j]%10;
}
count++;
printf("Case%d:
\n",count);
printf("%s+%s=",str1,str2);
flag=0;
for(i=1004;i>=0;i--)
if(flag||a[i])
{
printf("%d",a[i]);
flag=1;
}
printf("\n");
if(n!
=0)printf("\n");
}
return0;
}
三、1089A+BforInput-OutputPractice(I)
题目名称:
A+BforInput-OutputPractice(I)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
1089A+BforInput-OutputPractice(I)
ProblemDescription
YourtaskistoCalculatea+b.
Tooeasy?
!
Ofcourse!
Ispeciallydesignedtheproblemforacmbeginners.
Youmusthavefoundthatsomeproblemshavethesametitleswiththisone,yes,alltheseproblemsweredesignedforthesameaim.
Input
Theinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
15
1020
SampleOutput
6
30
参考答案
#include
intmain(void)
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF)
printf("%d\n",a+b);
return0;
}
四、1090A+BforInput-OutputPractice(II)
题目名称:
A+BforInput-OutputPractice(II)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
输入一开始就会说有n个InputBlock,下面接着是输入n个InputBlock。
参见:
HDOJ_1090(
1090A+BforInput-OutputPractice(II)
ProblemDescription
Yourtaskistocalculatea+b.
Input
InputcontainsanintegerNinthefirstline,andthenNlinesfollow.Eachlineconsistsofapairofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
Sampleinput
2
15
1020
Sampleoutput
6
30
参考答案
#include
intmain()
{
intn,i,a,b;
scanf("%d",&n);
for(i=0;i{
scanf("%d%d",&a,&b);
printf("%d\n",a+b);
}
return0;
}
五、1091A+BforInput-OutputPractice(III)
题目名称:
A+BforInput-OutputPractice(III)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
输入不说明有多少个InputBlock,但以某个特殊输入为结束标志。
参见:
HDOJ_1091(
1091A+BforInput-OutputPractice(III)
ProblemDescription
Yourtaskistocalculatea+b.
Input
Inputcontainsmultipletestcases.Eachtestcasecontainsapairofintegersaandb,onepairofintegersperline.Atestcasecontaining00terminatestheinputandthistestcaseisnottobeprocessed.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandbinoneline,andwithonelineofoutputforeachlineininput.
Sampleinput
15
1020
00
Sampleoutput
6
30
参考答案
#include
intmain()
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF&&(a!
=0||b!
=0))
printf("%d\n",a+b);
return0;
}
或
#include
intmain()
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF)
if(a==0&&b==0)
break;
else
printf("%d\n",a+b);
return0;
}
六、1092A+BforInput-OutputPractice(IV)
题目名称:
A+BforInput-OutputPractice(IV)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
以上几种情况的组合,可以参照如下网页。
参见:
HDOJ_1092(
1092A+BforInput-OutputPractice(IV)
ProblemDescription
YourtaskistoCalculatethesumofsomeintegers.
Input
Inputcontainsmultipletestcases.EachtestcasecontainsaintegerN,andthenNintegersfollowinthesameline.Atestcasestartingwith0terminatestheinputandthistestcaseisnottobeprocessed.
Output
Foreachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput.
SampleInput
41234
512345
0
SampleOutput
10
15
参考答案
#include
intmain()
{
intn,sum,i,t;
while(scanf("%d",&n)!
=EOF&&n!
=0)
{
sum=0;
for(i=0;i{
scanf("%d",&t);
sum=sum+t;
}
printf("%d\n",sum);
}
}
七、1093A+BforInput-OutputPractice(V)
题目名称:
A+BforInput-OutputPractice(V)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
以上几种情况的组合,可以参照如下网页。
参见:
HDOJ_1092(
1093A+BforInput-OutputPractice(V)
ProblemDescription
Yourtaskistocalculatethesumofsomeintegers.
Input
InputcontainsanintegerNinthefirstline,andthenNlinesfollow.EachlinestartswithaintegerM,andthenMintegersfollowinthesameline.
Output
Foreachgroupofinputintegersyoushouldoutputtheirsuminoneline,andwithonelineofoutputforeachlineininput.
SampleInput
2
41234
512345
SampleOutput
10
15
参考答案
#include
intmain()
{
intn,a,b,i,j,sum;
sum=0;
while(scanf("%d\n",&n)!
=EOF)
{
for(i=0;i{
scanf("%d",&b);
for(j=0;j
{
scanf("%d",&a);
sum+=a;
}
printf("%d\n",sum);
sum=0;
}
}
return0;
}
八、1094A+BforInput-OutputPractice(VI)
题目名称:
A+BforInput-OutputPractice(VI)
链接地址:
TimeLimit:
1Seconds MemoryLimit:
32768K
以上几种情况的组合,可以参照如下网页。
参见:
HDOJ_1092(
1094A+BforInput-OutputPractice(VI)
ProblemDescription
Yourtaskistocalculatethesumofsomeintegers.
Input
Inputcontainsmultipletestcases,andonecaseoneline.EachcasestartswithanintegerN,andthenNintegersfollowinthesameline.
Output
ForeachtestcaseyoushouldoutputthesumofNintegersinoneline,andwithonelineofoutputforeachlineininput.
SampleInput
41234
512345
SampleOutput
10
15
参考答案
#include
intmain()
{
intn,a,b,i,j,sum;
sum=0;
while(scanf("%d\n",&n)!
=EOF)
{
for(j=0;j{
scanf("%d",&a);
sum+=a;
}
printf("%d\n",sum);
sum=0;
}
return0;
}
输出
1、第一类输出
一个InputBlock对应一个OutputBlock,OutputBlock之间没有空行。
参见:
HDOJ_1089(
2、第二类输出
一个InputBlock对应一个OutputBlock,每个OutputBlock之后都有空行。
参见:
HDOJ_1095(
1095A+BforInput-OutputPractice(VII)
ProblemDescription
YourtaskistoCalculatea+b.
Input
Theinputwillconsistofaseriesofpairsofintegersaandb,separatedbyaspace,onepairofintegersperline.
Output
Foreachpairofinputintegersaandbyoushouldoutputthesumofaandb,andfollowedbyablankline.
SampleInput
15
1020
SampleOutput
6
30
参考答案
#include
intmain()
{
inta,b;
while(scanf("%d%d",&a,&b)!
=EOF)
printf("%d\n\n",a+b);
}
3、第三类输出
一个InputBlock对应一个OutputBlock,OutputBlock之间有空行。
参见:
HDOJ_1096(
1096A+BforInput-OutputPractice(VIII)
ProblemDescription
Yourtaskistocalculatethesumofsomeintegers.
Input
InputcontainsanintegerNinthefirstline,andthenNlinesfollow.EachlinestartswithaintegerM,andthenMintegersfollowinthesameline.
Output
Foreachgroupofinputinteg