zoukankan      html  css  js  c++  java
  • OpenJudge/Poj 1163 The Triangle

    1.链接地址:

    http://bailian.openjudge.cn/practice/1163

    http://poj.org/problem?id=1163

    2.题目:

    总时间限制:
    1000ms
    内存限制:
    65536kB
    描述
    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.
    输入
    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.
    输出
    Your program is to write to standard output. The highest sum is written as an integer.
    样例输入
    5
    7
    3 8
    8 1 0 
    2 7 4 4
    4 5 2 6 5
    样例输出
    30
    来源
    IOI 1994

    3.思路:

    DP

    4.代码:

     1 #include "stdio.h"
     2 //#include "stdlib.h"
     3 #define N 102
     4 int a[N][N];
     5 int main()
     6 {
     7     int count;
     8     int i,j;
     9     scanf("%d",&count);
    10     for(i = 0 ;i <count;i++)
    11     {
    12        for(j=0;j<=i;j++)
    13        {
    14            scanf("%d",&a[i][j]);
    15        }
    16     }
    17     for(i = count -2;i>=0;i--)
    18     {
    19         for(j=0;j<=i;j++)
    20         {
    21             a[i][j] += ((a[i+1][j]>a[i+1][j+1])?a[i+1][j]:a[i+1][j+1]);
    22         }
    23     }
    24     printf("%d",a[0][0]);
    25     //system("pause");
    26     return 1;
    27 } 
  • 相关阅读:
    jzoj 6278. 2019.8.5【NOIP提高组A】跳房子
    2019.08.05【NOIP提高组】模拟 A 组 总结
    HTML笔记
    html中的锚点设置
    前端HTML
    数据库设计(第一范式,第二范式,第三范式)
    MySQL之锁、事务、优化、OLAP、OLTP
    MySQL数据备份与还原(mysqldump)
    MySQl创建用户和授权
    MySQL之索引原理与慢查询优化
  • 原文地址:https://www.cnblogs.com/mobileliker/p/3572460.html
Copyright © 2011-2022 走看看