zoukankan      html  css  js  c++  java
  • nyoj 18 The Triangle



    The Triangle

    时间限制:1000 ms  |  内存限制:65535 KB
    难度:4
    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
    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    int main()
    {
        int n,maxx=0,num[105][105]= {0};
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=i; j++)
                scanf("%d",&num[i][j]);
        for(int i=2; i<=n; i++)
        {
            for(int j=1; j<=i; j++)
            {
                num[i][j]=num[i][j]+max(num[i-1][j],num[i-1][j-1]);
            }
        }
        for(int i=1; i<=n; i++)
            if(num[n][i]>maxx)
                maxx=num[n][i];
        printf("%d
    ",maxx);
        return 0;
    }
    //#include<stdio.h>
    //#include<string.h>
    //#include<algorithm>
    //using namespace std;
    //
    //int n,dp[110][110],map[110][110];
    //
    //int dfs(int i,int j)
    //{
    //    if(dp[i][j]!=-1)
    //        return dp[i][j];
    //    if(i==n) dp[i][j]=map[i][j];
    //    else
    //    {
    //        int x=dfs(i+1,j);
    //        int y=dfs(i+1,j+1);
    //        dp[i][j]=max(x,y)+map[i][j];
    //    }
    //    return dp[i][j];
    //}
    //int main()
    //{
    //    int i,j;
    //    scanf("%d",&n);
    //    for(i=1; i<=n; i++)
    //    {
    //        for(j=1; j<=i; j++)
    //        {
    //            scanf("%d",&map[i][j]);
    //            dp[i][j]=-1;
    //        }
    //    }
    //    printf("%d
    ",dfs(1,1));
    //    return 0;
    //}
    
    
    //#include <stdio.h>
    //#include <algorithm>
    //using namespace std;
    //int main()
    //{
    //    int n,num[105][105]= {0};
    //    scanf("%d",&n);
    //    for(int i=1; i<=n; i++)
    //        for(int j=1; j<=i; j++)
    //            scanf("%d",&num[i][j]);
    //    for(int i=n-1; i>=0; i--)
    //        for(int j=1; j<=i; j++)
    //            num[i][j]=num[i][j]+max(num[i+1][j],num[i+1][j+1]);
    //    printf("%d
    ",num[1][1]);
    //    return 0;
    //}
    //


    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    
    int n;
    int dp[110][110],map[110][110];
    
    int d(int i,int j)
    {
        if(dp[i][j]>=0) return dp[i][j];
        return dp[i][j]=map[i][j]+(i==n?0:(d(i+1,j)>d(i+1,j+1)?d(i+1,j):d(i+1,j+1)));
    }
    
    int main()
    {
        while(~scanf("%d",&n))
        {
            int i,j;
            for(i=1; i<=n; i++)
            {
                for(j=1; j<=i; j++)
                {
                    scanf("%d",&map[i][j]);
                    dp[i][j]=-1;
                }
            }
            int ans=d(1,1);
            printf("%d
    ",ans);
        }
        return 0;
    }


  • 相关阅读:
    父进程pid和子进程pid的大小关系
    static 和extern关键字
    linux源码下载
    tar命令
    USB开发——内核USB驱动+libusb开发方法
    microchip PIC芯片使用方法
    android下4G上网卡
    Modem常用概念
    4G上网卡NIDS拨号之Rmnet驱动
    Uboot源码解析
  • 原文地址:https://www.cnblogs.com/nyist-xsk/p/7264902.html
Copyright © 2011-2022 走看看