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  • POJ 2181

    直接DP,设当前第i个数为结尾的算奇数次和偶数次的得到的最大值。分开设即可。

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <climits>
    #include <string.h>
    #include <queue>
    #include <cmath>
    #include <vector>
    using namespace std;
    
    int main(){
    	int oddn,even,n,tmp,to,te;
    	while(scanf("%d",&n)!=EOF){
    		for(int i=1;i<=n;i++){
    			scanf("%d",&tmp);
    			if(i==1){
    				oddn=tmp;
    				even=0;
    			}
    			else{
    				to=even+tmp;
    				te=oddn-tmp;
    				oddn=max(oddn,to);
    				even=max(te,even);
    			}
    		}
    		printf("%d
    ",max(oddn,even));
    	}
    	return 0;
    }
    	
    

      

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  • 原文地址:https://www.cnblogs.com/jie-dcai/p/4300298.html
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