HDU_5586_Sum(最大连续子段和)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 756    Accepted Submission(s): 405

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)

 

题意：给出一个数字串 A1 ~ An，你可以选择一个连续的子串 [l, r]，使得里面所有的数字 Ai 的值变成 f(Ai) = (1890 * Ai + 143) % 10007，或者你也可以不要选择。最后要求所有数字的和最大是多少。

分析：最大连续子段和。很明显得可以想到，要使得最后的数字和最大，那么就是求一个区间 [l, r] 使得增量最大，也就是说使得 Σ (f(Ai) - Ai) 最大，那么很明显就转换成了最大连续子段和问题了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5586

代码清单：

/******************************************************************************* *** problem ID  : HDU_5586.cpp *** create time : Tue Dec 08 14:32:36 2015 *** author name : nndxy *** author blog : http://blog.csdn.net/jhgkjhg_ugtdk77 *** author motto: never loose enthusiasm for life, life is to keep on fighting! *******************************************************************************/#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <ctime>#include <vector>#include <cctype>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <bits/stdc++.h>using namespace std;#define exit() return 0#define setIn(name) freopen(name".in", "r", stdin)#define setOut(name) freopen(name".out", "w", stdout)#define debug(x) cout << #x << " = " << x << endltypedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;const int maxn = 100000 + 5;const int INF = 100000000 + 5;int n;int A[maxn];int B[maxn];int ans;void input() {ans = 0;for(int i = 1; i <= n; i++) {scanf("%d", &A[i]);ans += A[i];B[i] = (1890 * A[i] + 143) % 10007 - A[i];}}void solve() {int sum = 0, ret = 0;for(int i = 1; i <= n; i++) {if(sum + B[i] > 0) {sum += B[i];ret = max(ret, sum);}else sum = 0;}printf("%d\n", ans + ret);}int main() { while(scanf("%d", &n) != EOF) {input();solve();}   exit();}