zoukankan      html  css  js  c++  java
  • POJ 1163 The Triangle

    The Triangle
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 33476   Accepted: 19865

    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 diagonally down to the left or diagonally down to the right. 

    Input

    Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

    Output

    Your program is to write to standard output. The highest sum is written as an integer.

    Sample Input

    5
    7
    3 8
    8 1 0 
    2 7 4 4
    4 5 2 6 5

    Sample Output

    30
    #include <stdio.h>
    #include <iostream>
    #include <string.h>
    using namespace std;
    
    int maze[105][105];
    int dp[105][105];
    
    int main()
    {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= i; j++)
            {
                scanf("%d", &maze[i][j]);
            }
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= i; j++)
            {
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1]) + maze[i][j];
            }
        }
        int Max = -10000000;
        for (int i = 1; i <= n; i++)
        {
            Max = max(Max, dp[n][i]);    
        }
        printf("%d
    ", Max);
        return 0;
    }
  • 相关阅读:
    java 截取pdf
    webService 发送soap请求,并解析返回的soap报文
    常用网址
    扫描文件夹下代码行数
    CodeMIrror 简单使用
    常用 linux 命令(部分)
    windows下RabbitMQ 监控
    一定要写的日志
    创业思路
    10月9日后计划
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3207558.html
Copyright © 2011-2022 走看看