PAT甲级1096

1096. Consecutive Factors (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3*5*6*7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are supposed to find the maximum number of consecutive factors, and list the smallest sequence of the consecutive factors.

Input Specification:

Each input file contains one test case, which gives the integer N (1<N<2³¹).

Output Specification:

For each test case, print in the first line the maximum number of consecutive factors. Then in the second line, print the smallest sequence of the consecutive factors in the format "factor[1]*factor[2]*...*factor[k]", where the factors are listed in increasing order, and 1 is NOT included.

Sample Input:

630

Sample Output:

35*6*7

#include<cstdio>#include<vector>#include<cmath>#include<algorithm>using namespace std;int main(){int N;scanf("%d", &N);int i = 2, j = 2;int sqr = sqrt(1.0*N);int count = 0;int n = N; int max = 0, start = N;while (j <=sqr){i = j;while (i<=n){if (N%i == 0){N /= i;count++;i++;}elsebreak;}if (max< count){start = j;max = count;}N = n;//注意重新还原变量的值的语句应该放在合适位置，在分析过程时一定要注意哪些是一定会执行的，哪些用完后需要重新赋值//数据冲突类错误很多，明明是这次过程需要用的值但还是用了上次过程的值，就像容器用完一定要清空count = 0;j++;}if (max){printf("%d\n", max);for (int i = 0; i < max; i++){if (!i)printf("%d", start);elseprintf("*%d", start + i);}}else{printf("1\n");printf("%d", n);}return 0;}