ACM+输入输出入门..doc

HDOJ1089 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 ? Author lcy ? Recommend JGShining #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 输入格式：先输入有case数，再依次输入每个case 输出格式：一行一个结果 #include stdio.h 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 ? Author lcy ? Recommend JGShining 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 输入格式：每行输入一组case，当case中的数据满足某种情况时退出 输出格式：一行一个结果 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