NYOJ 18 The Triangle(简单dp)
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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
题意就是求从最上面一层到最下面一层路径数字相加的最大值,所以从下到上更新较为清楚
ac代码:
#include<stdio.h>#include<string.h>#include<stdlib.h>int main(){int a[100][1000];int i,j,n;scanf("%d",&n);memset(a,0,sizeof(a));for(i=0;i<n;i++){for(j=0;j<=i;j++){scanf("%d",&a[i][j]);}}for(i=n-2;i>=0;i--)//从下到上更新{for(j=0;j<i+1;j++){if(a[i+1][j]>a[i+1][j+1])a[i][j]+=a[i+1][j];elsea[i][j]+=a[i+1][j+1];}}printf("%d\n",a[0][0]);//最上面的就是最大值return 0;}
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