Trailing Zeroes (III)(lightoj 二分好题)

1138 - Trailing Zeroes (III)
Time Limit: 2 second(s) Memory Limit: 32 MB

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

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

Sample Input

Output for Sample Input

3

1

2

5

Case 1: 5

Case 2: 10

Case 3: impossible

题解:这道题要找末尾0的个数所需要的最小阶乘,因为想到10=2*5,然而2是很多数的质因数,数目要大于质因数5的数目,所以只考虑5的质因数的个数,便为末尾0的个数,所以

int q=0;
while(x){
q+=x/5;
x/=5;
}

这点便可以得出x!末尾0的个数;

用二分,从无穷大开始找,找到Q就是答案;

代码:

 #include<stdio.h>
const int MAXN=0x3f3f3f3f;//这个要足够大才能找到10^8
int getq(int x){
int q=;
while(x){
q+=x/;
x/=;
}
return q;
}
void erfen(int n){
int l=,r=MAXN;
while(l<=r){
int mid=(l+r)>>;
if(getq(mid)>=n)r=mid-;//二分这点注意
else l=mid+;
}
if(getq(l)==n)printf("%d\n",l);
else puts("impossible");
}
int main(){
int T,Q,flot=;
scanf("%d",&T);
while(T--){
scanf("%d",&Q);
printf("Case %d: ",++flot);
erfen(Q);
}
return ;
}
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