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hdu 5586 Sum 基础dp

2024-09-10 20:57:16   209人阅读

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)


Problem Description
There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod10007.After that,the sum of n numbers should be as much as possible.What is the maximum sum?
 

 

Input
There are multiple test cases.
First line of each case contains a single integer n.(1≤n≤105)
Next line contains n integers A1,A2....An.(0≤Ai≤104)
It‘s guaranteed that ∑n≤106.
 

 

Output
For each test case,output the answer in a line.
 

 

Sample Input
210000 999951 9999 1 9999 1
 

 

Sample Output
1999922033
 

 

Source
BestCoder Round #64 (div.2)
题目链接：点击传送
思路：基础dp；
　　　dp[i][0]表示dp前边没有改成f的总和
　　　dp[i][1]表示可以继续改成f的总和；
　　　dp[i][2]表示不可以继续改成f的总和；

dp[i][0]=dp[i-1][0]+a[i];
dp[i][1]=max(dp[i-1][1]+b[i],dp[i-1][0]+b[i]);
dp[i][2]=max(dp[i-1][2]+a[i],dp[i-1][1]+a[i]);

　　　b=f(a);

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<set>#include<map>using namespace std;#define ll long long#define pi (4*atan(1.0))#define eps 1e-14#define bug(x)  cout<<"bug"<<x<<endl;const int N=1e5+30010,M=1e6+10,inf=2147483647;const ll INF=1e18+10,mod=2147493647;ll a[N],b[N],dp[N][4];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]),b[i]=(1890*a[i]+143)%10007;        dp[0][0]=dp[0][1]=dp[0][2]=0;        for(int i=1;i<=n;i++)        {            dp[i][0]=dp[i-1][0]+a[i];            dp[i][1]=max(dp[i-1][1]+b[i],dp[i-1][0]+b[i]);            dp[i][2]=max(dp[i-1][2]+a[i],dp[i-1][1]+a[i]);        }        printf("%lld\n",max(dp[n][2],max(dp[n][0],dp[n][1])));    }    return 0;}

 

hdu 5586 Sum 基础dp


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