BC21&&hdoj5138&&hdoj5139&&hdoj5141 CET-6 test Formula LIS again

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概述

CET-6 test

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 613 Accepted Submission(s): 440

Problem Description

Small W will take part in CET-6 test which will be held on December

20th . In order to pass it he must remember a lot of words.
He remembers the words according to Ebbinghaus memory curve method.
He separates the words into many lists. Every day he picks up a new list, and in the next

1st,2nd,4th,7th,15th day, he reviews this list.
So every day he has many lists to review. However he is so busy, he does not know which list should be reviewed in a certain day. Now he invites you to write a program to tell him which list should to be reviewed in a certain day.
Lists are numbered from 1. For example list 1 should be reviewed in the

2nd,3rd,5th,8th,16th day.

Input

Multi test cases (about 100), every case contains an integer n in a single line.
Please process to the end of file.

[Technical Specification]

2≤n≤100000

Output

For each n，output the list which should be reviewed in the

nth day in ascending order.

Sample Input


2
100

Sample Output


1
85 93 96 98 99

Source

BestCoder Round #21

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=1010;
int num[10]={1,2,4,7,15};
int main()
{
int t,i,j,k,n;
while(scanf("%d",&n)!=EOF){
for(i=4;i>=1;--i){
if(n-num[i]>0)printf("%d ",n-num[i]);
}
printf("%dn",n-1);
}
return 0;
}

Formula

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1157 Accepted Submission(s): 399

Problem Description

f(n)=(∏i=1nin−i+1)%1000000007
You are expected to write a program to calculate f(n) when a certain n is given.

Input

Multi test cases (about 100000), every case contains an integer n in a single line.
Please process to the end of file.

[Technical Specification]

1≤n≤10000000

Output

For each n，output f(n) in a single line.

Sample Input


2
100

Sample Output


2
148277692

Source

BestCoder Round #21

规律好找但是打表超内存在线计算超时离线先保存数据在计算

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<list>
#include<cmath>
#include<vector>
#define MOD 1000000007
using namespace std;
const int maxn=100010;
int num[maxn];
struct Node{
int x,y,ans;
}A[maxn];
bool cmp(Node a,Node b){
return a.x<b.x;
}
void init(int n){
int num[5],cnt=0;num[0]=num[1]=1;
while(A[cnt].x==1){
A[cnt].ans=1;cnt++;
}
for(int i=2;i<=n;++i){
num[i%3]=(long long)num[(i-1)%3]*i%MOD;
num[(i-1)%3]=(long long)num[(i-2)%3]*num[i%3]%MOD;
while(i==A[cnt].x){
A[cnt].ans=num[(i-1)%3];cnt++;
}
}
}
bool cmp1(Node a,Node b){
return a.y<b.y;
}
int main()
{
int i,n=0;
while(scanf("%d",&A[n].x)!=EOF){
A[n].y=n;n++;
}
sort(A,A+n,cmp);
init(A[n-1].x);
sort(A,A+n,cmp1);
for(i=0;i<n;++i){
printf("%dn",A[i].ans);
}
return 0;
}

LIS again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 493 Accepted Submission(s): 184

Problem Description

A numeric sequence of ai is ordered if

a1<a2<…<aN . Let the subsequence of the given numeric sequence (

a1,a2,…,aN ) be any sequence (

ai1,ai2,…,aiK ), where

1≤i1<i2<…<iK≤N . For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and many others.
S[ i , j ] indicates (

ai,ai+1,ai+2,…,aj ) .
Your program, when given the numeric sequence (

a1,a2,…,aN ), must find the number of pair ( i, j) which makes the length of the longest ordered subsequence of S[ i , j ] equals to the length of the longest ordered subsequence of (

a1,a2,…,aN ).

Input

Multi test cases (about 100), every case occupies two lines, the first line contain n, then second line contain n numbers

a1,a2,…,aN separated by exact one space.
Process to the end of file.

[Technical Specification]

1≤n≤100000

0≤ai≤1000000000

Output

For each case，.output the answer in a single line.

Sample Input

Sample Output

1
3

Source

BestCoder Round #21

解题思路找出以i结尾的最长子序列长度等于总序列的长度并记录开头的最靠右的第一个元素即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=100010;
int dp[maxn][2],n;//dp[i][0]记录前i个的最长上升序列长度dp[i][1]记录该序列开头的第一个元素的位置
int la[maxn],lb[maxn],pos[maxn];
long long LIS(){
dp[1][0]=1;dp[1][1]=1;
memset(pos,0,sizeof(pos));//记录开头位置
lb[1]=la[1];int len=1;pos[1]=1;
for(int i=2;i<=n;++i){
int left=1,right=len;
if(la[i]>lb[len]){
len++;lb[len]=la[i];
pos[len]=max(pos[len],pos[len-1]);
dp[i][0]=len;dp[i][1]=pos[len];
continue;
}
while(left<=right){
int mid=(left+right)>>1;
if(la[i]>lb[mid]){
left=mid+1;
}
else {
right=mid-1;
}
}
lb[left]=la[i];
if(left==1)
pos[left]=i;
else
pos[left]=max(pos[left],pos[left-1]);
dp[i][0]=left;dp[i][1]=pos[left];
}
int l=0;
long long ans=0;
for(int i=1;i<=n;++i){
if(dp[i][0]==len){
l=max(l,dp[i][1]);
}
ans=ans+l;
}
return ans;
}
int main()
{
int i,j,k;
while(scanf("%d",&n)!=EOF){
for(i=1;i<=n;++i){
scanf("%d",&la[i]);
}
printf("%lldn",LIS());
}
return 0;
}