标签：his oss tin nbsp space otherwise algorithm img empty
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:
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.
Example
01011
0110
YES
0011
1110
NO
Note
In the first sample, the steps are as follows: 01011?→?1011?→?011?→?0110
题意：
对01串存在两个操作: (1) 从串首删除一个字符 (2) 在串尾添加一个字符（当1的个数是奇数时添加1，否则添加0）
题解：
不考虑0的作用，因为每次要么添加，要么删除，那么1的个数最多是原来的数或者原来的数加一。
1 #include<cmath> 2 #include<cstdio> 3 #include<string> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 9 int main() 10 { string a,b; 11 cin>>a>>b; 12 int suma=0,sumb=0; 13 for(int i=0;i<a.size();i++) if(a[i]==‘1‘) suma++; 14 for(int i=0;i<b.size();i++) if(b[i]==‘1‘) sumb++; 15 suma=suma+suma%2; 16 if(suma>=sumb) cout<<"YES"<<endl; 17 else cout<<"NO"<<endl; 18 }
标签：his oss tin nbsp space otherwise algorithm img empty
原文地址：http://www.cnblogs.com/zgglj-com/p/7260780.html
