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Rudolf and the Ball Game

seuleyang 03-15 08:00 阅读 4
c++算法

传送门

题意

思路

暴力枚举每一个妆台的转换条件

code

#include<iostream>
#include<cstdio>
#include<stack>
#include<vector>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<map>
#include<set>
#include<vector>
#define int long long
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define  long long
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int INF = 1e18 + 10;
const int N = 1e5 + 10;
const int M = 1e7 + 10;
const int mod = 1e9 + 7;
int n, m, k, ans;
int qcal(int a, int b) { int res = 1; while (b) { if (b & 1) res = res * a % mod; b >>= 1; a = a * a % mod; } return res; }
int a[N], b[N];
bool is_prime(int n) { if (n < 2) return false; for (int i = 2; i <= n / i; i++) { if (n % i == 0) { return false; } }return true; }
int f[1010][1010];
void gzy()
{
    
}
signed main()
{
    IOS;
    int _ = 1; cin >> _;
    while (_--) gzy();
    return 0;
}
/**
 *  ┏┓   ┏┓+ +
 * ┏┛┻━━━┛┻┓ + +
 * ┃       ┃
 * ┃   ━   ┃ ++ + + +
 *  ████━████+
 *  ◥██◤ ◥██◤ +
 * ┃   ┻   ┃
 * ┃       ┃ + +
 * ┗━┓   ┏━┛
 *   ┃   ┃ + + + +Code is far away from  
 *   ┃   ┃ + bug with the animal protecting
 *   ┃    ┗━━━┓ 神兽保佑,代码无bug 
 *   ┃  	    ┣┓
 *    ┃        ┏┛
 *     ┗┓┓┏━┳┓┏┛ + + + +
 *    ┃┫┫ ┃┫┫
 *    ┗┻┛ ┗┻┛+ + + +
 */

很可惜 这样T了

原因在每次都memset了1010*1010

题目说了n*m <= 2e5 所以我们应该手写 清零

#include<iostream>
#include<cstdio>
#include<stack>
#include<vector>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<map>
#include<set>
#include<vector>
#define int long long
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define  long long
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int INF = 1e18 + 10;
const int N = 1e5 + 10;
const int M = 1e7 + 10;
const int mod = 1e9 + 7;
int n, m, k, ans;
int qcal(int a, int b) { int res = 1; while (b) { if (b & 1) res = res * a % mod; b >>= 1; a = a * a % mod; } return res; }
int a[N], b[N];
bool is_prime(int n) { if (n < 2) return false; for (int i = 2; i <= n / i; i++) { if (n % i == 0) { return false; } }return true; }
int f[1010][1010];
void gzy()
{
    cin >> n >> m >> k;
    for(int i = 0;i <= m;i ++)
        for(int j = 0;j <= n;j ++)
            f[i][j] = 0;
    f[0][k] = 1;
    for(int i = 1;i <= m;i ++)
    {
        int tmp;
        char ch; cin >> tmp >> ch;
        for(int j = 1;j <= n;j ++)
        {
            if(ch == '0') 
            {
                if((j+n-tmp) % n == 0) f[i][j] = max(f[i][j],f[i-1][n]);
                else f[i][j] = max(f[i][j],f[i-1][(j+n-tmp) % n]);
            }
            else if(ch == '1')
            {
                if((j+tmp) % n == 0) f[i][j] = max(f[i][j],f[i-1][n]);
                else f[i][j] = max(f[i][j],f[i-1][(j+tmp) % n]);
            }
            else
            {
                if((j+n-tmp) % n == 0) f[i][j] = max(f[i][j],f[i-1][n]);
                else f[i][j] = max(f[i][j],f[i-1][(j+n-tmp) % n]);
                if((j+tmp) % n == 0) f[i][j] = max(f[i][j],f[i-1][n]);
                else f[i][j] = max(f[i][j],f[i-1][(j+tmp) % n]);
            }
        }
    }
    set<int> se;
    int cnt = 0;
    for(int i = 1;i <= n;i ++)
        if(f[m][i])
        {
             cnt ++;
             se.insert(i);
        }
    cout << cnt << endl;
    for(auto c:se) cout << c << ' ';
    cout << endl;
    
}
signed main()
{
    IOS;
    int _ = 1; cin >> _;
    while (_--) gzy();
    return 0;
}
/**
 *  ┏┓   ┏┓+ +
 * ┏┛┻━━━┛┻┓ + +
 * ┃       ┃
 * ┃   ━   ┃ ++ + + +
 *  ████━████+
 *  ◥██◤ ◥██◤ +
 * ┃   ┻   ┃
 * ┃       ┃ + +
 * ┗━┓   ┏━┛
 *   ┃   ┃ + + + +Code is far away from  
 *   ┃   ┃ + bug with the animal protecting
 *   ┃    ┗━━━┓ 神兽保佑,代码无bug 
 *   ┃  	    ┣┓
 *    ┃        ┏┛
 *     ┗┓┓┏━┳┓┏┛ + + + +
 *    ┃┫┫ ┃┫┫
 *    ┗┻┛ ┗┻┛+ + + +
 */
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