• HDU2069 Coin change HDU 1114 Piggy-Bank HDU1712 ACboy needs your help


    HDU 2069 Coin Change

    换零钱, 计算有多少方案数

    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    #define N 100010
    int dp[N], w[5], v[5], n, m;
    
    int main()
    {
        w[0] = 1; w[1] = 5; w[2] = 10; w[3] = 25; w[4] = 50;
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for(int i = 0; i < 5; i++)
    		{
    			for(int j = w[i]; j <= 7800; j++)
    			{
    				dp[j] = dp[j] + dp[j - w[i]];
    			}
    		}
    	while(~scanf("%d", &n))
    	{
    		printf("%d
    ", dp[n]);
    	}
    	return 0;
    }

    HDU 1114  Piggy-Bank

    这个是 求最小的  方案,而以前做的都是求最大

    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    #define N 100010
    int dp[N], w[N], v[N], n, m;
    
    int main()
    {
    	int t;
    	scanf("%d", &t);
    	while(t--)
    	{
    		int x, y, z;
    		scanf("%d %d", &x, &y);
    		z = y - x;
    		scanf("%d", &n);
    		for(int i = 0; i < n; i++)
    		{
    			scanf("%d %d", &v[i], &w[i]);
    		}
    		for(int i = 0;i <= z; i++)
    				dp[i] = 9999999;
    		dp[0] = 0;
    		for(int i = 0; i < n; i++)
    		{
    			for(int j = w[i]; j <= z; j++)
    			{
    				dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
    			}
    		}
    		if(dp[z] == 9999999 ) printf("This is impossible.
    ");
    		else printf("The minimum amount of money in the piggy-bank is %d.
    ", dp[z]);
    	}
    	return 0;
    }
    

    HDU1712 ACboy needs your help

    典型的分组背包,每个course 是一组, 注意三层循环的次序问题

    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    #define N 200
    int dp[N], w[N][N], n, m;
    
    int main()
    {
    	while(~scanf("%d %d", &n, &m))
    	{
    		if(m + n == 0) break;
    		memset(w, 0, sizeof(w));
    		for(int i = 1; i <= n; i++)
    		{
    			for(int j = 1; j <= m; j++)
    				scanf("%d", &w[i][j]);
    		}
    
    		memset(dp, 0, sizeof(dp));
    		for(int i = 1; i <= n; i++)
    		{
    			for(int j = m; j >= 0; j--)
    			{
    				for(int k = 0; k <= j; k++)
    					dp[j] = max(dp[j], dp[j - k] + w[i][k]);
    					//printf("dp[%d][%d] = %d
    ", i, j, dp[j]);
    			}
    		}
    		printf("%d
    ", dp[m]);
    	}
    	return 0;
    }
    



  • 相关阅读:
    FindWindowEx使用方法
    什么是VSync
    getParameter
    高等数学积分公式大全
    为什么没有好用的Android游戏引擎?
    Best Time to Buy and Sell Stock III
    5.3 适配器模式(4.1)
    在动态网络下实现分布式共享存储
    人类智商一般在多少左右?爱因斯坦的智商是多少?
    海量数据处理面试题集锦
  • 原文地址:https://www.cnblogs.com/tenlee/p/4420111.html
Copyright © 2020-2023  润新知