二分查找:在有序数组中,寻找特定数的的算法。
步骤:
- 先在判断当前数组 中间值mid(mid=left+(right-left)/2))是否等于所找数字n,若是,返回mid。
- 若不是,n>fx(mid),说明n在数组靠近right的部分,left=mid+1;n<mid,n在数组靠近left部分,right=mid-1;继续进行1中的判断。
- 进入循环的条件是:while(left<=right),若找不到,返回找不到的相关信息。
int binarysearch(int a,int b) { int left,right; while(left<=right) { int mid=low+(right-left)>>1; if(n<fx(mid)) right=mid-1; else if(n>fx(mid)) left=mid+1; else return mid; } return -1; }
例题:Trailing Zeroes (III)(LightOJ1138)
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
3
1
2
5
Sample Output
Case 1: 5
Case 2: 10
Case 3: impossible
题意:首先有T组测试数据,让你查找是否存在数字n,使n!末尾0的个数为Q。
解题思路:末尾为0,则需要2*5,在一个数中,2的个数远远多于5,则找到一个包含5的个数与Q相等数,这个数可能有好几个,找到最小的输出。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN=10000000000000;
ll f(ll n)//拆分min,看其包含多少个0
{
ll ans=0;
while(n) ans+=(n/=5);
return ans;
}
int main()
{
int t;ll Q;
scanf("%d",&t);
int kase=1;
while(t--)
{
scanf("%lld",&Q);
ll l=1,r=MAXN;
ll ans=0;
while(r>=l)//进行二分查找
{
ll mid=(l+r)>>1;
if(f(mid)==Q)
{
ans=mid;
r=mid-1;//找到后继续缩小范围,确保数是最小的
}
else if(f(mid)<Q) l=mid+1;
else r=mid-1;
}
printf("Case %d: ",kase++);
if(ans) printf("%lld\n",ans);
else printf("impossible\n");
}
return 0;
}