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and he define a function on the sequence -- 
, where 
is the length of the longest increasing subsequence ending with . 
in such a manner that equals to . Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one. is lexicographically smaller than sequence , if there is such number 
from 
to 
, that 
for 
and .
, indicating the number of test cases. For each test case: 
-- the length of the sequence. The second line contains integers 
. integers denoting the lexicographically smallest sequence. 3 1 10 5 5 4 3 2 1 3 1 3 5
1 1 1 1 1 1 1 2 3
英语题目很难度，但是题意很简单，给你一个整数序列，输出以每个数结尾的最长上升子序列的长度。
1 #include<cstdio> 2 #include<algorithm> 3 #define INF 0x3f3f3f3f 4 using namespace std; 5 int dp[100005]; 6 int a[100005]; 7 int b[100005]; 8 int main() 9 { 10 int t; 11 scanf("%d",&t); 12 while(t--) 13 { 14 int n,i,j; 15 scanf("%d",&n); 16 for(i = 1 ; i <= n ; i++) 17 { 18 scanf("%d",&a[i]); 19 dp[i]=0; 20 b[i]=INF; 21 } 22 for(i = 1 ; i <= n ; i++) 23 { 24 int k=lower_bound(b+1,b+n+1,a[i])-b; //函数返回值为大于a[i]的第一个值的位置，比如序列1 1 4 5，a[i]为2，返回值为3 25 dp[i]=k; //与二分法类似，将数放入新数组 26 b[k]=a[i]; 27 } 28 for(i = 1 ; i < n ; i++) 29 printf("%d ",dp[i]); 30 printf("%d\n",dp[n]); 31 } 32 }
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原文地址：http://www.cnblogs.com/yexiaozi/p/5765998.html
