杭电1019
程序代码:
Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105. Input Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer. Output For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer. Sample Input 2 3 5 7 15 6 4 10296 936 1287 792 1 Sample Output 105 10296 我的代码: #include <stdio.h> int main() { int a, b, temp, n, s[1000], t, i, m, j; scanf("%d", &m); while(m--) { scanf("%d", &n); for(i = 0; i < n; i++) scanf("%d", &s[i]); for(i = 0; i < n - 1; i++) for(j = 0; j < n - 1 - i; j ++) if(s[j] > s[j + 1]) { t = s[j]; s[j] = s[j + 1]; s[j + 1] = t; } if(s[n - 2] < s[n - 1]) { temp = s[n - 1]; s[n - 1] = s[n - 2]; s[n - 2] = temp; } a = s[n - 2]; b = s[n - 1]; while(b != 0) { temp = a % b; a = b; b = temp; } printf("%d\n", s[n - 1] / a * s[n - 2]); } return 0; } 地址: http://acm.hdu. 总是错误..怎么回事?