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

Sum

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

Problem 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

210000 999951 9999 1 9999 1

 

Sample Output

1999922033

分析：原数组之和加上增量数组的最大连续子区间之和
#include <iostream>#include <cstdio>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<n;i++){            scanf("%d",&c);            int r=(1890*c+143)%10007;            b[i]=r-c;            a[i]=c;            y+=c;        }        LL summ=0,maxm=0;        for ( int i = 0 ; i < n ; i++ ){           summ = max(0,summ)+b[i];           maxm = max(maxm,summ);        }        printf("%lld\n",y+maxm);    }    return 0;}

Array
http://acm.hdu.edu.cn/showproblem.php?pid=5587

Vicky 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.
Input
There 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)
Output
For each test case,output the answer in a line.

Sample Input
3
1
3
5

Sample Output
1
4
7

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 <iostream>#include <cstdio>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;}