数字三角形

Time Limit: 1 Sec  Memory Limit: 162 MB

题目连接

http://www.tyvj.cn/p/1044

Description

Input

Output

Sample Input

5 
7


3 8


8 1 0


2 7 4 4


4 5 2 6 5

Sample Output

HINT

题解：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 2001

#define mod 10007

#define eps 1e-9

//const int inf=0x7fffffff;   //无限大

const int inf=0x3f3f3f3f;

/*

*/

//**************************************************************************************

inline ll read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int dp[maxn][maxn];

int g[maxn][maxn];

int main()

{

    int n=read();

    for(int i=;i<=n;i++)

        for(int j=;j<=i;j++)

            g[i][j]=read();

    for(int j=;j<=n;j++)

        dp[n][j]=g[n][j];

    for(int i=n-;i>=;i--)

        for(int j=;j<=n;j++)

            dp[i][j]=max(dp[i+][j],dp[i+][j+])+g[i][j];

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

}