ZOJ 2725 SG函数 DP

给一个长度为6的有前导零的数字，每次可以使一位减少，最少减少1，最多减少到0，也可以在0右边的数字和0本身都删除，两人轮流操作，问先手是否必胜

一开始看错题，写了半天都不对，一种是SG函数的，发现SG函数vis数组范围是SG的选择，另一种是顺推的DP，DP要快很多

SG函数 1680ms

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef  long long LL;
const double PI = acos(-1.0);

template  inline  T MAX(T a, T b){if (a > b) return a;return b;}
template  inline  T MIN(T a, T b){if (a < b) return a;return b;}

const int N = 111;
const int M = 11111;
const LL MOD = 1000000007LL;
const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
const int INF = 0x3f3f3f3f;



int sg[10][10][10][10][10][10][10];

int cg(int l, int a[10])
{
    int b[10], i, j, k;
    int & ans = sg[l][a[0]][a[1]][a[2]][a[3]][a[4]][a[5]];
    if (l == 0)
    {
        ans = 0;
        return 0;
    }
    bool v[61];
    memset(v, false, sizeof(v));
    if (ans != -1) return ans;
    for (i = 0; i < l; ++i)
    {
        if (a[i] != 0)
        {
            for (j = 1; j <= a[i]; ++j)
            {
                a[i]-=j;
                k = cg(l, a);
                v[k] = 1;
                a[i]+=j;
            }
        }
        if (a[i] == 0)
        {
            for (j = i; j < l; ++j)
            {
                b[j] = a[j];
                a[j] = 0;
            }
            k = cg(i, a);
            v[k] = 1;
            for (j = i; j < l; ++j)
                a[j] = b[j];
        }
    }
    i = 0;
    while (v[i]) i++;
    ans = i;
//    printf("%d %d %d %d %d %d %d: %d\n", l, a[0], a[1], a[2], a[3], a[4],a[5], ans);
    return ans;
}



int main()
{
    char str[10];
    memset(sg, -1, sizeof(sg));
    while (scanf("%s", str) != EOF)
    {
        int l, num[6], i, j, k;
        l = strlen(str);

        memset(num, 0, sizeof(num));
        for (i = 0; i < l; ++i)
        {
            num[i] = str[i] - '0';
        }
        if (num[0] == 0) {printf("Yes\n"); continue;}
        if (cg(l, num) != 0)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

 

DP 30ms

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef  long long LL;
const double PI = acos(-1.0);

template  inline  T MAX(T a, T b){if (a > b) return a;return b;}
template  inline  T MIN(T a, T b){if (a < b) return a;return b;}

const int N = 111;
const int M = 11111;
const LL MOD = 1000000007LL;
const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
const int INF = 0x3f3f3f3f;


bool dp[1111111];

int callen(int x)
{
    int m = 0;
    while (x)
    {
        x /= 10;
        m++;
    }
    return m;
}



void solve(int x)
{
    int q, lim, m = x, l, bas = 1, i, j, k;
    l = callen(m);
    for (i = 0; i < l; ++i)
    {
        lim = (m / bas) % 10; q = m;
        for (j = 0; j < 9 - lim ; ++j)
        {
            q += bas;
            dp[q] = 1;
        }
        bas *= 10;
    }
    if (l < 6)
    {
        k = 6 - l; m = x;
        bas = 1;
        for (j = 0; j < k; ++j)
        {
            m *= 10;
            for (i = 0; i < bas; ++i)
                dp[m + i] = 1;
            bas *= 10;
        }
    }
}

void init()
{
    memset(dp, false, sizeof(dp));
    dp[0] = true;
    for (int i = 1; i <= 999999; ++i)
    {
        if (!dp[i]) solve(i);
    }
}




int main()
{
    char str[10];
    init();
    while (scanf("%s", str) != EOF)
    {
        int n, l;
        l = strlen(str);
        if (str[0] == '0')
        {
            printf("Yes\n");
            continue;
        }
        n = 0;
        for (int i = 0; i < l; ++i)
            n = n * 10 + str[i] - '0';
        if (dp[n]) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}