Factorial
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You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.
Input
One number Q written in the input (0<=Q<=10^8).
Output
Write "No solution", if there is no such number N, and N otherwise.
Sample Input
Sample test(s)
Input
2
Output
10
解题报告:
原来做过这样一个题,求n!末尾有几个0,先要把这个题搞懂。在n!要把每个数拆分成质因数的积,N!=2^a*3^b*5^c*7^d........因为只有2*5 才会出现 0,所以质因数有多少5,尾部就有多少个0,且2的个数会比5多很多,所以只要求5的个数即可。有一个公式:
设f(n)为n!末尾0的个数,则
0(0<n<5)
f(n!)=
k+f(k!)(k=n/5)
有了这个公式之后,关键代码就为:
while(n)
{
n/=5;
n/=5;
i+=n;
}
最后输出i的值即可。
同样,这个题
Q = N/5 + N/(5^2) + N/(5^3) + ...
由等比数列求和可得(设只到前k项):
Q = N(5^k - 1) / [4*(5^k)],由此得:
N = 4Q * [(5^k)/(5^k-1)]
注意:当Q为0时要输出1
#include <cstdio>#include <cstring>using namespace std; int fun(int x) { //求x尾部有多少个0 int sum = 0; while(x) { sum += x / 5; x /= 5; } return sum;} int main() { int Q; while(scanf("%d", &Q) != EOF) { if(Q == 0) { printf("1\n"); continue; } int N = Q * 4 / 5 * 5; //依据公式先大致算到4*Q(除5乘5是为了N正好是5的倍数),然后再往后推 while(fun(N) < Q) { N += 5; } if(fun(N) == Q) printf("%d\n", N); else printf("No solution\n"); } return 0;}
#include<iostream> #include<string.h> #include<stdio.h> #include<ctype.h> #include<algorithm> #include<stack> #include<queue> #include<set> #include<math.h> #include<vector> #include<map> #include<deque> #include<list> #define N 0x7fffffff using namespace std; int num0(int n) { int count=0; while(n) { count+=n/5; n/=5; } return count; } void find(int w) { int left=0,right=N,mid; while(left<=right) { mid=(left+right)>>1;//位运算速度快 // mid=(left+right)/2; if(num0(mid)>=w) right=mid-1; else left=mid+1; } if(num0(left)==w) printf("%d\n",left); else puts("No solution"); } int main() { int n; while(scanf("%d",&n)!=EOF) { if(n==0) printf("1\n"); else { find(n); } } return 0; }
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