hdu - 1087 - Super Jumping! Jumping! Jumping!

题意：求最大升序和。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1087

——>>设d[i]表示以第i个数为终点的最大升序和，然后从第1个数到第i-1个数为终点的最大升序和进行检查，向后递推即可。

#include <iostream>
#include <algorithm>using namespace std;int main()
{int N, a[1010], i, j;long long d[1010];      //d[i]表示以第i个数为终点的最大升序和while(cin>>N){if(!N) return 0;for(i = 1; i <= N; i++){cin>>a[i];d[i] = a[i];        //最小为其自己，无左拓展}long long max_sum = a[1];       //最大升序和存进max_sumfor(i = 2; i <= N; i++)     //第1个一定是它自己，故从第2个数开始for(j = 1; j < i; j++){if(a[j] < a[i]) d[i] = max(d[i], d[j]+a[i]);        //如果是升序，更新max_sum = max(max_sum, d[i]);       //更新}cout<<max_sum<<endl;}return 0;
}

今天用Java做了一次：

import java.util.Scanner;
public class Main {static final int maxn = 1000 + 10;public static void main(String[] args) {Scanner cin = new Scanner(System.in);int N, a[] = new int[maxn];long d[] = new long[maxn];while(cin.hasNextInt()){N = cin.nextInt();if(N == 0) break;for(int i = 1; i <= N; i++) d[i] = a[i] = cin.nextInt();long max = -1;for(int i = 1; i <= N; i++)for(int j = 1; j < i; j++){if(a[j] < a[i] && d[i] < d[j]+a[i]) d[i] = d[j]+a[i];if(d[i] > max) max = d[i];}System.out.println(max);}cin.close();}
}

转载于:https://www.cnblogs.com/xiaodanding/archive/2013/01/31/3266898.html

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