You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character “0” and “1”) a and b. Then you try to turn a into b using two types of operations:

Write parity(a) to the end of a. For example, .
Remove the first character of a. For example, . You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?

The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.

Input
The first line contains the string a and the second line contains the string b (1?≤?|a|,?|b|?≤?1000). Both strings contain only the characters “0” and “1”. Here |x| denotes the length of the string x.

Output
Print “YES” (without quotes) if it is possible to turn a into b, and “NO” (without quotes) otherwise.

Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011?→?1011?→?011?→?0110

唉，那個(gè)奇偶的地方真的讓人煩死了。直接上代碼

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxx=1e3+10; char s1[maxx]; char s2[maxx];int main() {while(cin>>s1>>s2){int len1=strlen(s1);int len2=strlen(s2);int sum1=0;int sum2=0;for(int i=0;i<len1;i++) if(s1[i]=='1') sum1++;for(int i=0;i<len2;i++) if(s2[i]=='1') sum2++;if(sum1%2) sum1+=1;if(sum1>=sum2) cout<<"YES"<<endl;else cout<<"NO"<<endl;}}

努力加油a啊，(^o)/~

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