南阳 18 The Triangle(dp)

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

573 88 1 0 2 7 4 44 5 2 6 5

样例输出

30

 #include<stdio.h>#include<string.h>#define maxn 100+5// #define max(a,b) a>b?a:bint max(int a,int b)  {  return a>b?a:b;   } int dp[maxn][maxn]; int main(){int n,i,j;scanf("%d",&n);memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){  for(j=1;j<=i;j++)     {  scanf("%d",&dp[i][j]);  }}for(i=n;i>=1;i--){   for(j=1;j<=i;j++)      {   dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];

                   /*求解释

                  当用注释掉的宏定义#define max(a,b) a>b?a:b;时，
               dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];
              
              dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];
             
            dp[i][j]=dp[i][j]+max(dp[i+1][j],dp[i+1][j+1]);
            为什么这三者运行结果都不一样呢？
                 */

      }
}
printf("%d\n",dp[1][1]);
     

return 0;
}