Sample OutputYES 1 1 Hint 样例中共有四堆石子,石子个数分别为7、6、9、3,每人每次可以从任何一堆石子中取出1个或者2个石子,小H有 必胜策略,事实上只要从第一堆石子中取一个石子即可。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#pragma GCC optimize(2)
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int n, m;
int a[maxn], b[maxn];
bool vis[maxn];
int sg[maxn];
void SG() {
for (int i = 1; i <= 2000; i++) {
ms(vis);
for (int j = 1; j <= m; j++) {
if (i - b[j] >= 0)vis[sg[i - b[j]]] = 1;
}
for (int j = 0; j <= 10; j++)
if (vis[j] == 0) {
sg[i] = j; break;
}
}
}
int main()
{
//ios::sync_with_stdio(0);
rdint(n);
for (int i = 1; i <= n; i++)rdint(a[i]);
rdint(m);
for (int i = 1; i <= m; i++)rdint(b[i]);
SG();
int ans = 0;
for (int i = 1; i <= n; i++)ans ^= sg[a[i]];
if (ans == 0)cout << "NO" << endl;
else {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i] - b[j] >= 0) {
if ((ans ^ (sg[a[i]]) ^ (sg[a[i] - b[j]])) == 0) {
cout << "YES" << endl;
cout << i << ‘ ‘ << b[j] << endl; return 0;
}
}
}
}
}
return 0;
}
原文:https://www.cnblogs.com/zxyqzy/p/10073430.html