CodeForces 755B PolandBall and Game（博弈）

题意：A和B两人每人都熟悉一些单词。A先开始，每人说一个单词，单词不能与两人之前说过的所有单词重复，谁无话可说谁输。两人可能有共同会的单词。

分析：因为要让对方尽量无单词可说，所以每个人优先说的都是两人共同会的单词，假设两人共同会的单词数为common。

A会的单词数为n，B会的单词数为m。

1、common若为偶数，则两人说完共同会的单词后，若n-common>m-common,则A赢。

2、common若为奇数，则两人说完共同会的单词后，若n-common>m-common-1,则A赢。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1000 + 10;
const int MAXT = 500 + 10;
using namespace std;
string a[MAXN];
string b[MAXN];
int main(){
    int n, m;
    while(scanf("%d%d", &n, &m) == 2){
        for(int i = 0; i < n; ++i){
           cin >> a[i];
        }
        for(int i = 0; i < m; ++i){
            cin >> b[i];
        }
        int common = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(a[i] == b[j]) ++common;
            }
        }
        if(common & 1){
            if(n > m - 1){
                printf("YES\n");
            }
            else printf("NO\n");
        }
        else{
            if(n > m){
                printf("YES\n");
            }
            else printf("NO\n");
        }
    }
    return 0;
}