**输入格式：有多个case输入，直到文件结束
输出格式：一行一个结果
Problem Description
Your 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.
Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.
Output
For 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 Input
1 5
10 20
Sample Output
6
30**
#include <stdio.h>
int main()
{
int a,b;
while( scanf( "%d%d" , &a , &b ) != EOF ) //输入直到文件结尾
{
printf( "%d\n" , a+b ); //一行一个结果
}
return 0;
}HDOJ1090
Input format: first enter the number of cases, and then input each case in turn.
Output format: one result per line
Problem Description
Your task is to Calculate a + b.
Input
Input contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.
Output
For 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 Input
2
1 5
10 20
Sample Output
6
30
#include <stdio.h>
int main()
{ int n,a,b;
scanf( "%d" , &n ); //输入的case数
while( n-- ) //控制输入
{ scanf( "%d%d" , &a , &b );
printf( "%d\n" , a+b ); //一行一个结果
}
return 0;
}HDOJ1091
Input format: Input a set of cases per line, and exit when the data in the case meets a certain condition
Output format: One line, one result
Problem Description
Your task is to Calculate a + b.
Input
Input 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.
Output
For 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 Input
1 5
10 20
0 0
Sample Output
6
30
Author
lcy
Recommend
JGShining
#include <stdio.h>
int main()
{
int a,b;
while( scanf( "%d%d" , &a , &b ) && (a||b) ) //输入直到满足a和b均为0结束
{
printf( "%d\n" , a+b ); //一行一个结果
}
return 0;
}HDOJ1092
input format: before each group of cases there is a number that controls the number of inputs, when the number is 0, the end
output format: one line of results
Problem Description
Your task is to Calculate the sum of some integers.
Input
Input 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.
Output
For each group of input integers you should output their sum in one line, and with one line of output for each line in input.
Sample Input
4 1 2 3 4
5 1 2 3 4 5
0
Sample Output
10
15
Author
lcy
Recommend
JGShining
#include <stdio.h>
int main()
{
int n,sum;
while( scanf( "%d" , &n ) && n ) //每组case前有一个控制该组输入数据的数，为0结束
{
int x;
sum = 0;
while( n-- ) //控制该组输入个数
{
scanf( "%d" , &x );
sum += x;
}
printf( "%d\n" , sum ); //一行一个结果
}
return 0;
}
HDOJ1093
Input format: At the beginning, there is a number that controls the total number of input cases, and each case has a number that controls the group of input data.
Output format: One line of results.
Problem Description
Your task is to calculate the sum of some integers.
Input
Input 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.
Output
For each group of input integers you should output their sum in one line , and with one line of output for each line in input.
Sample Input
2
4 1 2 3 4
5 1 2 3 4 5
Sample Output
10
15
Author
lcy
5
#include <stdio.h>
int main()
{
int casnum,n,sum;
scanf( "%d" , &casnum ); //控制总的输入case的数
while( casnum-- ) //控制总的输入个数
{
int x;
sum = 0;
scanf( "%d" , &n ); //每个case中控制该组输入个数
while( n-- )
{
scanf( "%d" , &x );
sum += x;
}
printf( "%d\n" , sum ); //一行一个结果
}
return 0;
}HDOJ1094
Input format: The total case is output to the end of the file. At the beginning of each case, a number to control the number of the group should be input.
Output format: One line of results.
Problem Description
Your task is to calculate the sum of some integers.
Input
Input 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.
Output
For each 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 Input
4 1 2 3 4
5 1 2 3 4 5
Sample Output
10
15
6
#include <stdio.h>
int main()
{
int n,sum;
while( scanf( "%d" , &n ) != EOF ) //输出到文件结尾
{
int x;
sum = 0;
while( n-- ) //控制该组输入个数
{
scanf( "%d" , &x );
sum += x;
}
printf( "%d\n" , sum ); //一行一个结果
}
return 0;
}HDOJ1095
Input format: Input until the end of the file
Output format: One result per line, and a blank line after the result is input
Problem Description
Your task is to Calculate a + b.
Input
The input will consist of a series of pairs of integers a and b , separated by a space, one pair of integers per line.
Output
For each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.
Sample Input
1 5
10 20
Sample Output
6
30
7
#include <stdio.h>
int main()
{
int a,b;
while( scanf( "%d%d" , &a , &b ) != EOF ) //输入直到文件结束
{
printf( "%d\n\n" , a+b ); //一行一个结果，结果输完后还有一个回车
}
return 0;
}HDOJ1096
Input format: Input the total number of cases at the beginning, and each group of cases has a number that controls the number of inputs in the group at the beginning.
Output format: One result per line, there is a carriage return between the two results, pay attention to the processing of the last case.
Problem Description
Your task is to calculate the sum of some integers.
Input
Input 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.
Output
For 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 Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3
Sample Output
10
15
6
#include <stdio.h>
int main()
{
int casnum,n,sum;
scanf( "%d" , &casnum ); //总的输入case数
while( casnum-- ) //控制输入组数
{
int x;
sum = 0;
scanf( "%d" , &n ); //控制每组的输入个数
while( n-- )
{
scanf( "%d" , &x );
sum += x;
}
printf( "%d\n" , sum ); //一行一个结果
if( casnum ) printf( "\n" ); //两两结果之间有一个回车，最后一个结果后面没有
}
return 0;
}