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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?
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.
For each test case,output the answer in a line.
2 10000 9999 5 1 9999 1 9999 1
19999 22033
思路：令Ai=f（Ai）-Ai，然后求一遍最大连续子序列和就能知道最多能增加的值。
一开始统计所有数的和ans，再加上最多增加的值就是答案了。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 100006 23 #define inf 1e12 24 int n; 25 int a[N]; 26 int A[N]; 27 int main() 28 { 29 while(scanf("%d",&n)==1){ 30 int ans=0; 31 for(int i=0;i<n;i++){ 32 scanf("%d",&a[i]); 33 ans+=a[i]; 34 } 35 for(int i=0;i<n;i++){ 36 A[i]=(a[i]*1890+143)%10007-a[i]; 37 //printf("===%d\n",A[i]); 38 } 39 int ThisSum=0,MaxSum=0; 40 for(int i=0;i<n;i++){ 41 ThisSum+=A[i]; 42 if(ThisSum>MaxSum){ 43 MaxSum=ThisSum; 44 }else if(ThisSum<0){ 45 ThisSum=0; 46 } 47 } 48 ans+=MaxSum; 49 printf("%d\n",ans); 50 51 } 52 return 0; 53 }
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原文地址：http://www.cnblogs.com/UniqueColor/p/5005119.html
