hdu 6047 Maximum Sequence (贪心)
来源:互联网 发布:超链接调用js方法 编辑:程序博客网 时间:2024/06/16 07:06
Maximum Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2325 Accepted Submission(s): 1091
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;
分析:贪心思想,将b数组升序排序后,此时第一次选的bi最小,在前n个a数组里能选到的数就最大,a数组往后加的第一个位置放的数肯定是最大的,所以,虽然每次从a中找数时,随着bi增大,坐标越来越靠后,但是最大的数已经筛选出来了,放在a[n+1]了,这符合贪心思想。
代码如下:
#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;const int MOD=1e9+7;int a[250005],b[250005];int mm[250005];int main(){ int n; long long ans; while(~scanf("%d",&n)) { ans=0; memset(mm,0,sizeof(mm)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); a[i]-=i; } for(int i=1;i<=n;i++) scanf("%d",&b[i]); sort(b,b+n); for(int i=n;i>0;i--) mm[i]=max(mm[i+1],a[i]); long long mx=(long long)mm[b[1]]; ans=(ans+mx)%MOD; mx-=n+1; for(int i=2;i<=n;i++) { long long tem=(long long)mm[b[i]]; tem=max(tem,mx); ans=(ans+tem)%MOD; } printf("%lld\n",ans); } return 0;}
阅读全文
0 0
- HDU 6047 Maximum Sequence (贪心)
- hdu 6047Maximum Sequence(贪心)
- HDU 6047 Maximum Sequence【贪心】
- hdu 6047 Maximum Sequence (贪心)
- HDU 6047 Maximum Sequence (贪心,线段树)
- Hdu 6047 Maximum Sequence【贪心+优先队列】
- HDU 6047 Maximum Sequence 数论 贪心
- HDU 6047 Maximum Sequence (贪心)
- HDU 6047 Maximum Sequence(贪心)
- HDU 6047 Maximum Sequence 贪心 区间最值
- [HDU]-6047 Maximum Sequence
- hdu 6047 Maximum Sequence
- [HDU 6047]Maximum Sequence
- HDU 6047 Maximum Sequence
- hdu 6047 Maximum Sequence
- HDU 6047 Maximum Sequence
- hdu 6047 Maximum Sequence
- HDU 6047 Maximum Sequence
- Linux进程剖析
- C++基础面试题总结
- HDU 3342 Legal or Not
- 使用SQL Server链接服务器访问MySQL的一些坑
- 树莓派远程桌面配置
- hdu 6047 Maximum Sequence (贪心)
- JavaScript代码编写分享到边侧栏
- Subarray Sum
- 1701H2 王建瑜 2017.10.14 连续第四天总结
- 工具
- <深度学习系列>基于numpy和python的反向传播算法的实现与分析
- 考虑实现Comparable接口。
- 最近 5 年 133 个 Java 面试问题列表(下)
- HDU 2647 Reward