The Battle of Chibi
So there is only one way left for Yu Zhou, send someone to fake surrender Cao Cao. Gai Huang was selected for this important mission. However, Cao Cao was not easy to believe others, so Gai Huang must leak some important information to Cao Cao before surrendering.
Yu Zhou discussed with Gai Huang and worked out NN information to be leaked, in happening order. Each of the information was estimated to has aiai value in Cao Cao's opinion.
Actually, if you leak information with strict increasing value could accelerate making Cao Cao believe you. So Gai Huang decided to leak exact MM information with strict increasing value in happening order. In other words, Gai Huang will not change the order of the NN information and just select MM of them. Find out how many ways Gai Huang could do this.
InputThe first line of the input gives the number of test cases, T(1≤100)T(1≤100). TT test cases follow.
Each test case begins with two numbers N(1≤N≤103)N(1≤N≤103) and M(1≤M≤N)M(1≤M≤N), indicating the number of information and number of information Gai Huang will select. Then NN numbers in a line, the ithith number ai(1≤ai≤109)ai(1≤ai≤109) indicates the value in Cao Cao's opinion of the ithith information in happening order.OutputFor each test case, output one line containing Case #x: y, where xx is the test case number (starting from 1) and yy is the ways Gai Huang can select the information.
The result is too large, and you need to output the result mod by 1000000007(109+7)1000000007(109+7).Sample Input
2 3 2 1 2 3 3 2 3 2 1
Sample Output
Case #1: 3
Case #2: 0
Hint
In the first cases, Gai Huang need to leak 2 information out of 3. He could leak any 2 information as all the information value are in increasing order. In the second cases, Gai Huang has no choice as selecting any 2 information is not in increasing order.
题意：求一个序列中长度为m的最长上升子序列个数。
设dp[i][j]表示以当前位置i结束的长度为j的序列个数。
容易想到O(n^3)的做法。开始想用记忆化搜索将复杂度降低一些，再加上若干剪枝，仍然会T。
考虑到问题涉及小于（大于）某个数的数的和，因此应该想到用树状数组优化。
使用lower_bound将所有数离散化，还要注意在dp的同时更新树状数组，时间复杂度为O(n^2*logn)。
//注释部分为优化前
#include<bits/stdc++.h> #define MAX 1005 #define MOD 1000000007 #define lowbit(x) x&(-x) using namespace std; typedef long long ll; int a[MAX],b[MAX]; int n,m; ll dp[MAX][MAX]; ll sum(int x,int y){ ll ans=0; while(1<=x){ ans+=dp[x][y]; ans%=MOD; x-=lowbit(x); } return ans; } void add(int x,int y,int z){ while(x<=n){ dp[x][y]+=z; dp[x][y]%=MOD; x+=lowbit(x); } } int main() { int tt=0,t,i,j,k; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); for(i=1;i<=n;i++){ scanf("%d",&a[i]); b[i]=a[i]; } sort(b+1,b+n+1); for(i=1;i<=n;i++){ a[i]=lower_bound(b+1,b+n+1,a[i])-b; } memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++){ //dp[i][1]=1; add(a[i],1,1); for(j=2;j<=m;j++){ //dp[i][j]=dp[i-1][j]; //for(k=1;k<i;k++){ // if(a[k]<a[i]){ // dp[i][j]+=dp[k][j-1]; // dp[i][j]%=MOD; // } //} ll temp=sum(a[i]-1,j-1); add(a[i],j,temp); } } //printf("Case #%d: %I64d\n",++tt,dp[n][m]); printf("Case #%d: %I64d\n",++tt,sum(n,m)); } return 0; }