思路

博弈论模板题.
直接像动态规划一样求出所有的函数值并异或起来.
这里需要输出方案,枚举和,取完后若函数值为,那么就是合法方案.
复杂度为(表示).

代码

#include<bits/stdc++.h>
using namespace std;
#define i64 long long
#define fp( i, b, e ) for ( int i(b), I(e); i <= I; ++i )
#define fd( i, b, e ) for ( int i(b), I(e); i >= I; --i )
#define go( i, b ) for ( int i(b), v(to[i]); i; v = to[i = nxt[i]] )
template<typename T> inline void cmax( T &x, T y ){ x < y ? x = y : x; }
template<typename T> inline void cmin( T &x, T y ){ y < x ? x = y : x; }
#define getchar() ( p1 == p2 && ( p1 = bf, p2 = bf + fread( bf, 1, 1 << 21, stdin ), p1 == p2 ) ? EOF : *p1++ )
char bf[1 << 21], *p1(bf), *p2(bf);
template<typename T>
inline void read( T &x ){ char t(getchar()), flg(0); x = 0;
    for ( ; !isdigit(t); t = getchar() ) flg = t == '-';
    for ( ; isdigit(t); t = getchar() ) x = x * 10 + ( t & 15 );
    flg ? x = -x : x;
}

clock_t t_bg, t_ed;
int N, M, n, a[15], b[15];
int sg[1005], d[1005];

signed main(){
    t_bg = clock();
    read(N);
    fp( i, 1, N ) read(a[i]), cmax(n, a[i]);
    read(M);
    fp( i, 1, M ) read(b[i]);
    sg[0] = 0;
    fp( i, 1, n ){
        fp( j, 1, M ) if ( i >= b[j] )
            d[sg[i - b[j]]] = i;
        else break;
        fp( j, 0, M ) if ( d[j] != i ){ sg[i] = j; break; }
    } int ans(0);
    fp( i, 1, N ) ans ^= sg[a[i]];
    if ( ans ){
        printf("YES\n");
        for ( int i = 1; i <= N; ++i ){
            int t(ans ^ sg[a[i]]);
            fp( j, 1, M ) if ( a[i] >= b[j] && sg[a[i] - b[j]] == t )
                return printf( "%d %d\n", i, b[j] ), 0;
        }
    }else printf("NO\n");
    t_ed = clock();
    fprintf( stderr, "\n========info========\ntime : %.3f\n====================\n", (double)( t_ed - t_bg ) / CLOCKS_PER_SEC );
    return 0;
}