• BestCoder Round #70


    模拟 1001 Jam's math problem

    判断b ^ 2 - 4ac是否为完全平方数.当delta < 0, sqrt (delta) 输出为nan, 但是好像也能计算?

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <string>
    #include <iostream>
    #include <queue>
    #include <map>
    #include <cmath>
    
    typedef long long ll;
    const int N = 1e5 + 5;
    const int INF = 0x3f3f3f3f;
    
    int main(void)	{
    	int T;	std::cin >> T;
    	while (T--)	{
    		ll a, b, c;	std::cin >> a >> b >> c;
    		ll d = b * b - 4 * 1ll * a * c;
    		if (d < 0)	puts ("NO");
    		else	{
    			ll e = sqrt (d);
    			if (e * e == d)	puts ("YES");
    			else	puts ("NO");
    		}
    	}
    
    	return 0;
    }
    /*
    long long	long double
    sqrt () HDU %I64d
    */
    

    01DP 1002 Jam's balance

    原来是背包,暴力枚举是不可做的.dp[i][j] 表示前i个砝码,能称多少的重量,当然砝码也能放在另一边也就是j - w,如果j - w < 0, -(j - w)表示将物品放在堆满砝码的一边,w放在另一边.

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <string>
    #include <queue>
    #include <map>
    #include <iostream>
    #include <cmath>
    
    int dp[22][2002];
    
    int main(void)	{
    	int T;	scanf ("%d", &T);
    	while (T--)	{
    		int n;	scanf ("%d", &n);
    		memset (dp, 0, sizeof (dp));
    		dp[0][0] = 1;
    		for (int w, i=1; i<=n; ++i)	{
    			scanf ("%d", &w);
    			for (int j=0; j<=2000; ++j)	{
    				if (!dp[i-1][j])	continue;
    				dp[i][j] = dp[i][j+w] = 1;
    				dp[i][abs (j-w)] = 1;
    			}
    		}
    		int m;	scanf ("%d", &m);
    		for (int w, i=1; i<=m; ++i)	{
    			scanf ("%d", &w);
    			if (dp[n][w])	puts ("YES");
    			else	puts ("NO");
    		}
    	}
    	return 0;
    }
    //01dp "twice"
    

      

    dp(优化) 1003 Jam's maze

    这种问方案数的以后应该要想到可能是DP.f[x1][y1][x1][y1]=f[x1][y1-1][x2][y2+1]+f[x1][y1-1][x2+1][y2]+f[x1-1][y1][x2][y2+1]+f[x1-1][y1][x2+1][y2].对它优化,f[i][x1][x2],通过走的步数以及x能计算出当前走到的y,再优化点把第一维改成滚动的,即dp[now][x1][x2] <- dp[now^1][x1][x2] ... dp[now^1][x1][x2]是(x1, y1 - 1) 和 (x2, y2 - 1), 其他类似

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <string>
    #include <queue>
    #include <map>
    #include <cmath>
    
    typedef long long ll;
    const int N = 5e2 + 5;
    const int MOD = 5201314;
    int dp[2][N][N];
    char str[N][N];
    
    void add(int &a, int b)	{
    	a += b;
    	if (a >= MOD)	a %= MOD;
    }
    
    int main(void)	{
    	int T;	scanf ("%d", &T);
    	while (T--)	{
    		int n;	scanf ("%d", &n);
    		for (int i=1; i<=n; ++i)	scanf ("%s", str[i] + 1);
    		if (str[1][1] != str[n][n])	puts ("0");
    		else	{
    			memset (dp, 0, sizeof (dp));
    			int now = 0;
    			dp[now][1][n] = 1;
    			for (int s=1; s<n; ++s)	{
    				now ^= 1;	memset (dp[now], 0, sizeof (dp[now]));
    				for (int x1=1; x1<=n; ++x1)	{
    					for (int x2=n; x2>=1; --x2)	{
    						if (x1 - 1 > s || n - x2 > s)	continue;
    						int y1 = 1 + s - (x1 - 1);
    						int y2 = n - (s - (n - x2));
    						if (str[x1][y1] != str[x2][y2])	continue;
    						add (dp[now][x1][x2], dp[now^1][x1][x2]);
    						add (dp[now][x1][x2], dp[now^1][x1][x2+1]);
    						add (dp[now][x1][x2], dp[now^1][x1-1][x2]);
    						add (dp[now][x1][x2], dp[now^1][x1-1][x2+1]);
    					}
    				}
    			}
    			int ans = 0;
    			for (int i=1; i<=n; ++i)	add (ans, dp[now][i][i]);
    			printf ("%d
    ", ans);
    		}
    	}
    
    	return 0;
    }
    

      

    编译人生,运行世界!
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  • 原文地址:https://www.cnblogs.com/Running-Time/p/5181007.html
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