BestCoder Round #64 (div.2) 1002 and 1003

BestCoder Round #64 (div.2) 1002 and 1003

Sum

​​Description

There is a number sequenceA1,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

2 10000 9999 5 1 9999 1 9999 1

Sample Output

19999 22033

分析：原数组之和加上增量数组的最大连续子区间之和

#include #include using namespace std;const int N=1e5+10;int a[N];int b[N];typedef long long LL;LL sum[N];LL max(LL a,LL b){ if(a>b) return a; else return b;}int main() //(1890x+143)mod10007{ //freopen("cout.txt","w",stdout); int n,c; while(cin>>n){ LL y=0; for(int i=0;i

Array​​​is a magician who loves math. She has great power in copying and creating.One day she gets an array {1}。 After that, every day she copies all the numbers in the arrays she has, and puts them into the tail of the array, with a signle '0' to separat.Vicky wants to make difference. So every number which is made today (include the 0) will be plused by one.Vicky wonders after 100 days, what is the sum of the first M numbers.InputThere are multiple test cases.First line contains a single integer T, means the number of test cases.(1≤T≤2∗103)Next T line contains, each line contains one interger M. (1≤M≤1016)OutputFor each test case,output the answer in a line.Sample Input3135Sample Output147

学好递归是很有必要的！

0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5……

红色区域第四层：

数字：1 2 3 4

个数：1 3 3 1

第五层：

数字：1 2 3 4 5

个数：1 4 6 4 1

数字个数和杨辉三角相关，前N+1层数字一共有个，10^16小于1<<50，那么用递归的话while循环求出各层数字和加上递归调用最坏结果：50+X*50 可以满足时间限制。

#include #include using namespace std;typedef long long LL;LL work(LL m){ LL ans=1,cnt=1; while(1>0){ if(cnt==m) return ans; if(cnt==m-1) return ans+1; if(2*cnt+1>m){ ans=ans+m-cnt; ans=ans+work(m-cnt-1); return ans; } ans=2*ans+cnt+1; cnt=2*cnt+1; }}int main(){ int t; cin>>t; LL m; while(t--){ scanf("%I64d",&m); printf("%I64d\n",work(m)); } return 0;}

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