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简单的DP题,就是LIS变形而来的LIS和,注意并不要求数之间要相邻

 

LIS解析详见:http://blog.csdn.net/niteip/article/details/7444973

 

#include<stdio.h>
#include<string.h>

int a[1005],DP[1005];

int max(int a,int b){
	return a>b?a:b;
}

int main(){
	int n,re;
	while(~scanf("%d",&n) && n){
		int i,j;
		re=-99999999;
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		DP[0]=a[0];
		for(i=1;i<n;i++){
			DP[i]=a[i];
			for(j=0;j<i;j++){
				if(a[j]<a[i]){
					if(DP[i]<DP[j]+a[i])
						DP[i]=DP[j]+a[i];
				}
			}
		}
		for(i=0;i<n;i++){
			if(re<DP[i])
				re=DP[i];
		}
		printf("%d\n",re);
	}
	return 0;
}


 

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