完整word版杭电ACM试题答案

【杭电ACM1000】
A +
B Problem
Problem Description
Calculate A + B.
Input
Each line will contain two integers A and B. Process to end of file.
Output
For each case, output A + B in one line.
Sample Input
1 1
Sample Output
2
# include <stdio.h>
int main()
{
int a, b;
while(scanf(%d%d, &a, &b)!=EOF)
printf(%d\n, a+b);
return 0;
}
【杭电ACM1001】
Sum Problem
Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
Input
The input will consist of a series of integers n, one integer per line.
Output
For each case, 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 Input
1 100
Sample Output
1 5050
# include <stdio.h>
int main()
{
int n, i, sum = 0;
while(scanf(%d, &n)!=EOF)
{
for(i=1; i<=n; ++i)
sum = sum + i;
printf(%d\n\n, sum);
sum = 0;
}
return 0;
}
【杭电ACM1002】
A +
B Problem II
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 = 1111111111111111110
#include<stdio.h>
#include<string.h>
int shu(char a)
{
return (a-'0');
}
int main(){
char a[1000],b[1000];
int num[1001];
int n,i,j=1,al,bl,k,t;
scanf(%d,&n);
while(n--)
{
getchar();
if(j!=1)
printf(\
);
scanf(%s,a);
al=strlen(a);
scanf(%s,b);
bl=strlen(b);
k=(al>bl)?al:bl;
for(i=0;i<=k;i++)
num[i]=0;
t=k;
for(k;al>0&&bl>0;k--)
{
num[k]+=shu(a[--al])+shu(b[--bl]);
if(num[k]/10)
{
num[k-1]++;
num[k]%=10;
}
}
while(al>0)
{
num[k--]+=shu(a[--al]);
if(num[k+1]/10)
{
num[k]++;
num[k+1]%=10;
}
}
while(bl>0)
{
num[k--]+=shu(b[--bl]);
if(num[k+1]/10)
{
num[k]++;
num[k+1]%=10;
}
}
printf(Case %d:\n,j++);
printf(%s + %s = ,a,b);
for(i=0;i<=t;i++)
{
if(i==0&&num[i]==0)
i++;
printf(%d,num[i]);
}
printf(\
);
}
return 0;
}。