Trailing Zeroes (III)(lightoj 二分好题)

1138 - Trailing Zeroes (III)

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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=0;
    while(x){
        q+=x/5;
        x/=5;
    }
    return q;
}
void erfen(int n){
    int l=0,r=MAXN;
    while(l<=r){
        int mid=(l+r)>>1;
        if(getq(mid)>=n)r=mid-1;//二分这点注意 
        else l=mid+1;
    }
    if(getq(l)==n)printf("%d\n",l);
    else puts("impossible");
}
int main(){
    int T,Q,flot=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&Q);
        printf("Case %d: ",++flot);
         erfen(Q);
    }
    return 0;
}