2016研究生推免上机考试（校外）

题目网址：

http://bailian.openjudge.cn/tm201602/

A, 石头剪刀布, AC

#include
using namespace std;

const int MAX = 105;

int x_a[MAX];
int x_b[MAX];
int N, NA, NB;

int main()
{
    cin >> N >> NA >> NB;
    for(int i=0; i> x_a[i];
    for(int i=0; i> x_b[i];
    int a_beat_b_num = 0;
    int b_beat_a_num = 0;
    for(int i=0; i b_beat_a_num)
        cout << "A" << endl;
    else if(a_beat_b_num == b_beat_a_num)
        cout << "draw" << endl;
    else
        cout << "B" << endl;
    return 0;
}

B, 字符串判等， AC

#include
using namespace std;
char to_lower(char A)
{
    if(A >= 'A' && A <='Z')
    {
        return A + ('a'-'A');
    }
    return A;
}
int main()
{
    char A;
    string s1, s2;
    getline(cin, s1);
    getline(cin, s2);
    int l1 = s1.length();
    int l2 = s2.length();
    int p1 = 0;
    int p2 = 0;
    int flag = 1;
    while(p1

C, 图像旋转, AC

#include
using namespace std;

int a[105][105];
int main()
{
    int m, n;
    cin >> m >> n;
    for(int i=0; i> a[i][j];
    for(int j=0; j=0; i--)
        {
            if(i == m-1)
                cout << a[i][j];
            else
                cout << " " << a[i][j];
        }
        cout << endl;
    }
    return 0;

}

D,棋盘问题，AC*

• 自己想复杂了。

#include
#include
using namespace std;

const int MAX = 9;
char grad[MAX][MAX];
int flag[MAX][MAX];

int ans;
int n, k;
void dfs(int find_num, int x, int y)
{
    //cout << endl;
    //cout << "before find_num: " << find_num << endl;
    //cout << x << " " << y << endl;
    if(find_num == k)
    {
        ans++;
        return;
    }
    for(int i=x; i> n >> k)
    {
        if(n == -1 && k == -1)
            return 0;
        for(int i=0; i> grad[i][j];
        ans = 0;
        memset(flag, -1, sizeof(flag));
        dfs(0,0,0);
        cout << ans << endl;
    }
    return 0;
}

• 应该按行搜
  https://www.cnblogs.com/Asimple/p/5528323.html

E,Divisibility, WA

不明白自己代码为什么Wa了

#include
using namespace std;

const int MAX = 10000+1;
const int MAXSUM = 10000*10000+4;
int a[MAX];
bool dp[2][MAXSUM];
int N, K;
int main()
{
    cin >> N >> K;
    int M = 0;
    for(int i=1; i<=N; i++){
        cin >> a[i];
        if(a[i] < 0)
        {
            a[i] = - a[i];
        }
        M += a[i];
    }
    for(int i=0; i<=M; i++)
        dp[0][i] = false;
    dp[0][0] = true;
    for(int i=1; i<=N; i++)
    {
        for(int j=0; j<=M; j++)
        {
            if((j-a[i]>=0 && dp[(i-1)&1][j-a[i]]) || dp[(i-1)&1][j+a[i]])
                dp[i&1][j] = true;
            else
                dp[i&1][j] = false;
        }
    }
    int flag = 0;
    for(int i=0; i<=M; i++)
    {
        if(dp[N&1][i] && i % K == 0)
        {
            flag = 1;
            break;
        }
    }
    if(flag)
        cout << "Divisible" << endl;
    else
        cout << "Not divisible" << endl;

    return 0;
}

AC 代码：

#include
#include
#include
using namespace std;
int n,k;
int a[10010];
bool dp[10010][110];// dp[i][j] = true 表示处理到第 i 个数对 k 取余为 j 是成立的 
int main()
{
	while(~scanf("%d %d",&n,&k))
	{
		memset(dp,false,sizeof(dp));
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		dp[1][( (a[1] % k) + k ) % k]=true;// 这里用了同余定理 (a+b)%c=(a%c+b%c)%c,a[1]对 k 取余为 ( (a[1] % k) + k ) % k是显然成立的 
		for(int i=2;i<=n;i++)
		{
			for(int j=0;j

F, 画家问题， AC(挑战程序设计P153)

#include
#include
using namespace std;

const int MAX = 17;
const int INF = MAX * MAX + 100;
int flip[MAX][MAX];
int base[MAX][MAX];
int n;
int dir[4][2] = {{0,1},{-1,0},{0,-1},{0,0}};
int is_white(int x, int y)
{
    int flip_num = 0;
    for(int i=0; i<4; i++)
    {
        int nx = x + dir[i][0];
        int ny = y + dir[i][1];
        if(nx < 0 || nx >= n || ny < 0 || ny >= n)
            continue;
        flip_num += flip[nx][ny];
    }
    if((base[x][y] + flip_num) % 2)
        return 1;
    return 0;
}
int main()
{
    cin >> n;
    char ch;
    for(int i=0; i> ch;
            if(ch == 'w')
                base[i][j] = 1;
            else
                base[i][j] = 0;
        }
    }

    int ans = INF;
    for(int s=0; s<(1<> i) & 1;
        }
        for(int r=1; r