George is a cat, so he really likes to play. Most of all he likes to play with his array of positive integers b. During the game, George modifies the array by using special changes. Let's mark George's current array as b1, b2, ..., b|b| (record |b| denotes the current length of the array). Then one change is a sequence of actions:
- Choose two distinct indexes i and j (1 ≤ i, j ≤ |b|; i ≠ j), such that bi ≥ bj.
- Get number v = concat(bi, bj), where concat(x, y) is a number obtained by adding number y to the end of the decimal record of number x. For example, concat(500, 10) = 50010, concat(2, 2) = 22.
- Add number v to the end of the array. The length of the array will increase by one.
- Remove from the array numbers with indexes i and j. The length of the array will decrease by two, and elements of the array will become re-numbered from 1 to current length of the array.
George played for a long time with his array b and received from array b an array consisting of exactly one number p. Now George wants to know: what is the maximum number of elements array b could contain originally? Help him find this number. Note that originally the array could contain only positive integers.
InputThe first line of the input contains a single integer p (1 ≤ p < 10100000). It is guaranteed that number p doesn't contain any leading zeroes.
Print an integer — the maximum number of elements array b could contain originally.
9555
4
10000000005
2
800101
3
45
1
1000000000000001223300003342220044555
17
19992000
1
310200
2
Let's consider the test examples:
- Originally array b can be equal to {5, 9, 5, 5}. The sequence of George's changes could have been: {5, 9, 5, 5} → {5, 5, 95} → {95, 55} → {9555}.
- Originally array b could be equal to {1000000000, 5}. Please note that the array b cannot contain zeros.
- Originally array b could be equal to {800, 10, 1}.
- Originally array b could be equal to {45}. It cannot be equal to {4, 5}, because George can get only array {54} from this array in one operation.
Note that the numbers can be very large.
贪心，递归找出最小的一个数，然后再把较大的数进行分解。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
string s;
int re =1;
void erfen(int beg,int end);
int main()
{
ios::sync_with_stdio(false);
cin >> s;
int beg = 0 ;
int end = s.size()-1;
erfen(beg,end);
cout << re<<endl;
return 0;
}
void erfen(int beg,int end)
{
if(beg >= end)
return ;
for(int i=end-1;i>=beg+(beg+end)/2;i--)
{
if(s[i+1]=='0')
continue;
// cout << i<<endl;
if(i-beg+1==end-i)
{
string s1;
for(int j=beg;j<=i;j++)
{
s1+=s[j];
}
s1+='\0';
string s2;
for(int j=i+1;j<=end;j++)
{
s2 +=s[j];
}
s2+='\0';
if(s1>=s2)
{
//cout << s1<<endl<<s2<<endl;
re++;
erfen(beg,i);
erfen(i+1,end);
break;
}else
continue;
}
re++;
erfen(beg,i);
erfen(i+1,end);
break;
}
}