经典的DP问题,DP思想也很直接: 直接贴代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ; int n, a[max_size][max_size]; ][max_size]; void initiate(){ memset(a,-,sizeof(a)); memset(f,,sizeof(f)); ;i&l…

Description 73 88 1 02 7 4 44 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally…

The Triangle Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36918   Accepted: 22117 Description 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed…

The Triangle Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 59698   Accepted: 35792 Description 73 88 1 02 7 4 44 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44997   Accepted: 27174 Description 73 88 1 02 7 4 44 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that…

一.Description 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diago…

经典的数塔模型. 动态转移方程:  dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+p[i][j]; #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <algorithm> #include <stri…

1. 题目描述给定几个三角形拼成一个百慕大三角形. 2. 基本思路基本思路肯定是搜索,关键点是剪枝.(1) 若存在长度为$l$的边,则一定可以拼成长度为$k \cdot l$的三角形,则可拼成长度为$k \cdot l$的百慕大三角形:(2) 长度超过百慕大三角形的边长的三角形没有任何价值:(3) 百慕大三角形中每个正三角形可以作为正多边形的顶点,倒三角形可以作为倒正三角形的顶点.因此,可以将每个三角形映射到二维坐标,高度为$y$,倒三角形的$x$坐标为偶数,正三角形的$x$坐标为奇数.对于每个…

/* 倒三角形(Triangle) 输入正整数n<=20,输出一个n层的倒等腰三角形. 0 ######### 9 = 2* n-1 1 ####### 7 = 2*(n-1)-1 2 ##### 5 = 2*(n-2)-1 3 ### 3 = 2*(n-3)-1 4 # 1 = 2*(n-4)-1 */ #include <iostream> using namespace std; void nString(char ch, int num){ while(num > 0){…

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [], [,4], [6,,7], [4,,8,3] ] The minimum path sum from top to bottom is 11 (i.e.,…