1096 Consecutive Factors (20 分)

1096 Consecutive Factors (20 分)

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<2​31​​).

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:

3 5*6*7

 思路：

N不会被除自己以外大于根号N的整数整除，因此遍历只需要从2到根号N即可。

用 long long 防止 int 溢出

设置变量temp和j

找到数目最多的序列，如果有多个则输出最小的序列，即序列第一个数最小的。

 

#include <cstdio> #include <cmath> typedef long long LL; int main(){ LL n; scanf("%lld",&n); LL sqr=(LL)sqrt(n); LL anslen=0,ans1=0; for(LL i=2;i<=sqr;i++){ LL temp=1;//temp的位置必须在for循环中，不然在while中被改变后再break就会出现错误 LL j=i; while(1){ temp=temp*j; if(n%temp!=0)break; if(j-i+1>anslen){ anslen=j-i+1; ans1=i; } j++; } } if(anslen==0)printf("1\n%lld",n); else{ printf("%lld\n",anslen); for(LL i=0;i<anslen;i++){ printf("%lld",ans1+i); if(i<anslen-1)printf("*"); } } return 0; }