PAT甲级真题-1007 Maximum Subsequence Sum详解优化

1007 Maximum Subsequence Sum

题目链接

https://pintia.cn/problem-sets/994805342720868352/problems/994805514284679168

解题思路

1.一开始的思路好像是叫“在线处理法”；但这种方法能过大部分是因为题目规定了全为负值时取0的特殊情况；而且，无法过值为非正数的情况；

for(int i=1;i<=n;i++){if(sum+Num[i]<0){sum=0;lnum=0;}else sum+=Num[i],lnum+=1;if(ans<sum){r=i;l=r-lnum+1;ans=sum;}}

2.另外一种思路就是dp，用dp[i]记录以i结尾的最大子序和，(动规NB);
用pre[i]记录子序的起始下标；一开始没想到这种方法是因为钻了牛角尖，没意识到只需要求出所有以i结尾的最大子序和，然后比较就可以了；

 for(int i=2;i<=n;i++){if(dp[i-1]+Num[i]>Num[i]){dp[i]=dp[i-1]+Num[i];pre[i]=pre[i-1];}else {dp[i]=Num[i];pre[i]=i;}if(ans<dp[i]){ans=dp[i];ansi=i;}}

总结

想dp解法没头绪的时候，就用暴力试一试；暴力出奇迹；

代码展示

#include<bits/stdc++.h>
using namespace std;
/*
10
-10 1 2 3 4 -5 -23 3 7 -21
6
-2 11 -4 13 -5 -2
*/
int Num[10005];
int dp[10005];//以第i项结尾的最大子序和
int pre[10005];
int main(){int n;cin>>n;bool flag=false;for(int i=1;i<=n;i++){cin>>Num[i];if(Num[i])flag=true;}dp[1]=Num[1];pre[1]=1;int ans=-1,ansi;for(int i=2;i<=n;i++){if(dp[i-1]+Num[i]>Num[i]){dp[i]=dp[i-1]+Num[i];pre[i]=pre[i-1];}else {dp[i]=Num[i];pre[i]=i;}if(ans<dp[i]){ans=dp[i];ansi=i;}}if(ans==-1)cout<<0<<" "<<Num[1]<<" "<<Num[n];else cout<<ans<<" "<<Num[pre[ansi]]<<" "<<Num[ansi];return 0;
}

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PAT甲级真题-1007 Maximum Subsequence Sum详解优化

1007 Maximum Subsequence Sum

题目链接

解题思路

总结

代码展示

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