Codeforces 387C George and Number【思维】
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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.
The 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.
题目大意:
给出一个数,问我们可以拆分成多少个数,使得两两组合的过程中,让大的数放在前边能够构成原数。(参考一下Note就行);
思路:
我们从前向后遍历,贪心的将带有0的部分给前边的数,然后判定一下两个数合并是否能够合法即可。
Ac代码:
#include<stdio.h>#include<string.h>using namespace std;char a[1500000];int main(){ while(~scanf("%s",a)) { int ans=0; int n=strlen(a); int j; for(int i=0;i<n;i=j) { for(j=i+1;j<n;j++) { if(a[j]!='0')break; } if(j-i>i||j-i==i&&a[i]>a[0])ans=1; else ans++; } printf("%d\n",ans); }}
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