杭电暑期多校集训—Maximum Sequence
来源:互联网 发布:多玩数据库 7.0 编辑:程序博客网 时间:2024/05/20 23:39
Maximum Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1887 Accepted Submission(s): 671
Problem Description
Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.
Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}:an+1…a2n . Just like always, there are some restrictions on an+1…a2n : for each number ai , you must choose a number bk from {bi}, and it must satisfy ai ≤max{aj -j│bk ≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai } modulo 109 +7 .
Now Steph finds it too hard to solve the problem, please help him.
Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}:
Now Steph finds it too hard to solve the problem, please help him.
Input
The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.
Output
For each test case, print the answer on one line: max{∑2nn+1ai } modulo 109 +7。
Sample Input
48 11 8 53 1 4 2
Sample Output
27HintFor the first sample:1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;
题意:根据公式ai≤max{aj-j│bk≤j<i},对数列{ai},{bi},进行处理
用样例来分析
a: 8 11 8 5
a[i]-i: 7 9 5 1
b:3 1 4 2
先将b排序得到 b: 1 2 3 4
那么取第一个b[0]=1 那么从a1开始找,取最大值9
a:8 11 8 5 9
a[i]-i: 7 9 5 1 4
b[1]=2 从a2开始找,还是取到 9
a: 8 11 8 5 9 9
a[i]-i: 7 9 5 1 4 3
后面继续这样取最终得到的是
a: 8 11 8 5 9 9 5 4 多出来的就是 27
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;int a[3000020];int b[250005];int c[230005];int main(){ long long int n; int ds=0; int Mo = 1e9 + 7; while(~scanf("%lld",&n)) { long long int ans=0,sum=0; ds++; if(ds>20) break; for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) scanf("%d",&b[i]); sort(b+1,b+n+1); for(int i=n;i>0;i--) { if(a[i]-i>=ans) ans=a[i]-i; c[i]=ans; } int max1=c[1]-n-1; for(int i=1;i<=n;i++) { if(c[b[i]]>max1) a[n+i]=c[b[i]]; else a[n+i]=max1; if(a[n+i]-n-i>max1) max1=a[n+i]-n-i; sum+=a[n+i]; if(sum>=Mo) sum%=Mo; } printf("%lld\n",sum); } return 0;}
阅读全文
0 0
- 杭电暑期多校集训—Maximum Sequence
- 杭电暑期多校集训—KazaQ's Socks
- 杭电暑期多校集训—Is Derek lying?
- 杭电暑期多校集训—Regular polygon
- 杭电暑期多校集训—RXD and math
- 杭电暑期多校集训—RXD's date
- 杭电暑期多校集训—Questionnaire
- 杭电暑期多校集训—RXD and dividing
- 杭电暑期多校集训—Euler theorem
- 杭电暑期多校集训—Kolakoski
- 杭电暑期多校集训— hard challenge
- 杭电暑期多校集训—Killer Names
- 2009ACM多校联合暑期集训(4)——杭电专场
- 杭电暑期多校集训—Time To Get Up
- 杭电暑期集训-Add More Zero
- HDU6047 Maximum Sequence(贪心,暑期训练1003)
- 2009ACM多校联合暑期集训(1)——TJU专场
- 2009ACM多校联合暑期集训(2)——TJU专场
- RecyclerView 夜间模式
- (贪心, 字符串, 位运算)Codeforces Round #402 E. Bitwise Formula
- 解决myeclipse2017ci7破解后闪退问题
- xtrabackup新版详细说明(version-2.4.1)
- codeforces 463B Caisa and Pylons
- 杭电暑期多校集训—Maximum Sequence
- JavaScript 风格指南(2)
- maven启动jetty 加载jar包
- oracle 12c企业版 收费标准(2017.7月)
- 自定义一个函数区分数组或者对象
- 某日下⑤于钟楼4302lab
- laravel 基础教程 —— 请求
- HDU 6055 Regular polygon 暴力枚举
- Java中this关键字的四种用法详解(含代码举例)