hdoj 5586 Sum 【dp】

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 60    Accepted Submission(s): 40

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

 

题意：给定n个元素的序列a[]，f[i] = (1890*a[i]+143) % 10007，你可以选择改变任意一个连续的区间（或者不改变），使区间里面的a[]变成f[]，问这个序列的最大元素和。

思路：求出f[]-a[]的代价v[]，然后找出最大连续子序列和就行了。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (100000+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;LL F(LL x){    return (1890 * x + 143) % 10007;}LL a[MAXN], f[MAXN];int main(){    int n;    while(Ri(n) != EOF)    {        LL ans = 0;        for(int i = 0; i < n; i++)        {            Rl(a[i]); ans += a[i];            f[i] = F(a[i]) - a[i];        }        LL sum = 0, s = 0;        for(int i = 0; i < n; i++)        {            s += f[i];            if(sum < s)                sum = s;            if(s < 0)                s = 0;        }        Pl(ans + sum);    }    return 0;}